You are given three strings a, b and c of the same length n. The strings consist of lowercase English letters only. The i-th letter of a is a_i, the i-th letter of b is b_i, the i-th letter of c is c_i.

For every i (1 ≤ i ≤ n) you must swap (i.e. exchange) c_i with either a_i or b_i. So in total you'll perform exactly n swap operations, each of them either c_i ↔ a_i or c_i ↔ b_i (i iterates over all integers between 1 and n, inclusive).

For example, if a is "code", b is "true", and c is "help", you can make c equal to "crue" taking the 1-st and the 4-th letters from a and the others from b. In this way a becomes "hodp" and b becomes "tele".

Is it possible that after these swaps the string a becomes exactly the same as the string b?

Input

The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a string of lowercase English letters a.

The second line of each test case contains a string of lowercase English letters b.

The third line of each test case contains a string of lowercase English letters c.

It is guaranteed that in each test case these three strings are non-empty and have the same length, which is not exceeding 100.

Output

Print t lines with answers for all test cases. For each test case:

If it is possible to make string a equal to string b print "YES" (without quotes), otherwise print "NO" (without quotes).

You can print either lowercase or uppercase letters in the answers.

Example

Input


4
aaa
bbb
ccc
abc
bca
bca
aabb
bbaa
baba
imi
mii
iim


Output


NO
YES
YES
NO

Note

In the first test case, it is impossible to do the swaps so that string a becomes exactly the same as string b.

In the second test case, you should swap c_i with a_i for all possible i. After the swaps a becomes "bca", b becomes "bca" and c becomes "abc". Here the strings a and b are equal.

In the third test case, you should swap c_1 with a_1, c_2 with b_2, c_3 with b_3 and c_4 with a_4. Then string a becomes "baba", string b becomes "baba" and string c becomes "abab". Here the strings a and b are equal.

In the fourth test case, it is impossible to do the swaps so that string a becomes exactly the same as string b.
import java.io.*;
import java.math.*;
import java.util.*;

/**
 *
 * @author Saju
 *
 */

public class Main {

	private static int dx[] = { 1, 0, -1, 0};
	private static int dy[] = { 0, -1, 0, 1};

	private static final long INF = (long) (1e15);
	private static final int INT_INF = Integer.MAX_VALUE;
	private static final long NEG_INF = Long.MIN_VALUE;
	private static final int NEG_INT_INF = Integer.MIN_VALUE;
	private static final double EPSILON = 1e-10;

	private static final int MAX = 1007;
	private static final long MOD = 1000000007;
	
	private static final int MAXN = 100007;
	private static final int MAXA = 10000009;
	private static final int MAXLOG = 22;

	public static void main(String[] args) {

		InputReader in = new InputReader(System.in);
		PrintWriter out = new PrintWriter(System.out);
		

/*



*/
		
		int test = in.nextInt();
		for (int t = 1; t <= test; t++) {
			char[] a = in.next().toCharArray();
			char[] b = in.next().toCharArray();
			char[] c = in.next().toCharArray();
			
			int len = a.length;
			boolean ok = true;
			
			for(int i = 0; i < len; i++) {
				if(a[i] != c[i] && b[i] != c[i] ) {
					ok = false;
					break;
				}
			}
			
			if(ok) {
				out.println("YES");
			}
			else {
				out.println("NO");
			}
		
		}
		
		
		out.flush();
		out.close();
		System.exit(0);
	}
	
	
	private static class Graph{
		private int node;
		private List<Edge>[] adj;
		
		Graph(int node){
			this.node = node;
			adj = new ArrayList[this.node];
			for(int i = 0; i < node; i++) {
				adj[i] = new ArrayList<Main.Edge>();
			}
		}
		
		void addEdge(int u, int v, long cost) {
			Edge e = new Edge(u, v, cost);
			Edge e1 = new Edge(v, u, cost);
			adj[u].add(e);
			adj[v].add(e1);
		}
		
		long[] calculateShortestPath(int source) {
			
			long dist[] = new long[node];
			PriorityQueue<Node> queue = new PriorityQueue<Main.Node>(new Comparator<Node>() {

				@Override
				public int compare(Node o1, Node o2) {
					int val = Long.compare(o1.distance, o2.distance);
					return val;
					
				}
			});
			
			Arrays.fill(dist, INT_INF);
			
			queue.add(new Node(source, 0));
			dist[source] = 0;
			
			while(!queue.isEmpty()) {
				Node nn = queue.remove();
				int u = nn.node;
				long udist = nn.distance;
				if(udist > dist[u]) {
					continue;
				}
				for(Edge e : adj[u]) {
						int v = e.v;
						long vdist = e.cost;
						long arrive = udist + vdist;
						if(arrive < dist[v]) {
							dist[v] = arrive;
							queue.add(new Node(v, dist[v]));
//							System.out.println(u + " " + v + " " + vdist + " " + dist[v]);
						}
					
				}
				
			}
			return dist;
			
		}
	}
	
	private static class Node{
		int operation;
		int node;
		long distance;
		Node(int operation, int node, long distance){
			this.operation = operation;
			this.node = node;
			this.distance = distance;
		}
		
		Node(int node, long distance){
			this.node = node;
			this.distance = distance;
		}
	}
	
	private static class Edge{
		int u;
		int v;
		long cost;
		
		Edge(int u, int v, long cost){
			this.u = u;
			this.v = v;
			this.cost = cost;
		}
		
		int getNeighbourIndex(int node) {
			if(this.u == node) {
				return v;
			}
			return u;
		}
	}
	
	private static boolean isPalindrome(String str) {
		StringBuilder sb = new StringBuilder();
		sb.append(str);
		String str1 = sb.reverse().toString();
		return str.equals(str1);
 	}

	private static String reverseString(String str) {
		StringBuilder sb = new StringBuilder();
		sb.append(str);
		return sb.reverse().toString();
	}
	
	
	private static double distance(Point p1, Point p2) {
		long divx = p2.x - p1.x;
		long divy = p2.y - p1.y;
		divx = divx * divx;
		divy = divy * divy;
		return Math.sqrt(divx + divy);
	}
	
	private static String getBinaryStr(int n, int j) {
		String str = Integer.toBinaryString(n);
		int k = str.length();
		for(int i = 1; i <= j - k; i++) {
			str = "0" + str;
		}
		
		return str;
	}

	// O(log(max(A,M)))
	static long modInverse(long a, long m) {
		extendedEuclid(a, m);
		return (x % m + m) % m;
	}

	static long x;
	static long y;
	static long gcdx;

	private static void extendedEuclid(long a, long b) {
		if (b == 0) {
			gcdx = a;
			x = 1;
			y = 0;
		} else {
			extendedEuclid(b, a % b);
			long temp = x;
			x = y;
			y = temp - ((a / b) * y);
		}
	}
	
	private static void generatePrime(int n) {
		 //O(NloglogN)
		boolean arr[] = new boolean[n + 5];
		Arrays.fill(arr, true);
		for(int i = 2; i * i <=n; i++){
			if(arr[i] == true){
				for(int j = i * i; j <= n; j+=i){
					arr[j] = false; 
				}
			}
		}
		int count = 0;
		int start = 0;
		for(int i = 2; i <= n; i++){
			if(arr[i] == true){
//				System.out.println(i + " ");
				count++;
			}
			if(count == (start * 100) + 1) {
//				System.out.println(i);
				start++;
			}
		}
		System.out.println();
		System.out.println(count);
		
	}
	
	private static Map<Long, Long> primeFactorization(long n, long m){
		Map<Long, Long> map = new HashMap<>();
		
		for(long i = 2; i <= Math.sqrt(n); i++){
			if(n % i == 0){
			long count = 0;
			while(n % i == 0){
				count++;
				n = n / i;
			}
			long val = count * m;
			map.put(i, val);
//			System.out.println("i: " + i + ", count: " + count);
		}
			
		}
		if(n != 1) {
//			System.out.println(n);
			map.put(n, m);
		}
		return map;
	}

	private static class Pair<T>{
		T a;
		T b;
		Pair(T a, T b){
			this.a = a;
			this.b = b;
		}
	}
	
	private static class SegmentTree{
		int n;
		private final int MAXN = 100007;
		private long[] tree = new long[MAXN << 2];
		private long[] lazy = new long[MAXN << 2];
		private long[] arr = new long[MAXN];
		
		
		/***
		 * 
		 * arr is 1 based index.
		 * 
		 * @param arr
		 * 
		 */
		SegmentTree(int n, long[] arr){
			this.n = n;
			for(int i = 0; i < arr.length; i++) {
				this.arr[i] = arr[i];
			}
		}
		
		void build(int index, int left, int right) {
			if(left == right) {
				tree[index] = arr[left];
			}
			else {
				int mid = (left + right) / 2;
				
				build(index * 2, left, mid);
				build((index * 2) + 1, mid + 1, right);
				tree[index] = max(tree[(index * 2)], tree[(index * 2) + 1]);
			}
		}
		
		long query(int node, int left, int right, int start, int end) {
			
			if(left > end || right < start) {
				return NEG_INF;
			}
			
			if(left >= start && right <= end) {
				return tree[node];
			}
			int mid = (left + right) / 2;
			long val1 = query(2 * node, left, mid, start, end);
			long val2 = query(2 * node + 1, mid + 1, right, start, end);
			
			return max(val1, val2);
		}
		
		void update(int node, int left, int right, int idx, long val) {
			if(left == right) {
				tree[node] += val;
			}
			else {
				int mid = (left + right) / 2;
				if(idx <= mid) {
					update(2 * node, left, mid, idx, val);
				}
				else {
					update(2 * node + 1, mid + 1, right, idx, val);
				}
				tree[node] = max(tree[(2 * node) + 1], tree[(2 * node)]);
			}
		}
		
		

		 void updateRange(int node, int start, int end, int l, int r, long val)
		{
		    if(lazy[node] != 0)
		    { 
		        // This node needs to be updated
		        tree[node] =  lazy[node];    // Update it
		        if(start != end)
		        {
		            lazy[node*2] = lazy[node];                  // Mark child as lazy
		            lazy[node*2+1] = lazy[node];                // Mark child as lazy
		        }
		        lazy[node] = 0;                                  // Reset it
		    }
		    if(start > end || start > r || end < l)              // Current segment is not within range [l, r]
		        return;
		    if(start >= l && end <= r)
		    {
		        // Segment is fully within range
		        tree[node] =  val;
		        if(start != end)
		        {
		            // Not leaf node
		            lazy[node*2] = val;
		            lazy[node*2+1] = val;
		        }
		        return;
		    }
		    int mid = (start + end) / 2;
		    updateRange(node*2, start, mid, l, r, val);        // Updating left child
		    updateRange(node*2 + 1, mid + 1, end, l, r, val);   // Updating right child
		    tree[node] = max(tree[node*2], tree[node*2+1]);        // Updating root with max value 
		}

		 long queryRange(int node, int start, int end, int l, int r)
		{
		    if(start > end || start > r || end < l)
		        return 0;         // Out of range
		    if(lazy[node] != 0)
		    {
		        // This node needs to be updated
		        tree[node] =  lazy[node];            // Update it
		        if(start != end)
		        {
		            lazy[node*2] = lazy[node];         // Mark child as lazy
		            lazy[node*2+1] = lazy[node];    // Mark child as lazy
		        }
		        lazy[node] = 0;                 // Reset it
		    }
		    if(start >= l && end <= r)             // Current segment is totally within range [l, r]
		        return tree[node];
		    int mid = (start + end) / 2;
		    long p1 = queryRange(node*2, start, mid, l, r);         // Query left child
		    long p2 = queryRange(node*2 + 1, mid + 1, end, l, r); // Query right child
		    return max(p1, p2);
		}
		
		 void buildRange(int node, int low, int high){
			if(low == high){
				tree[node] =  arr[low];
				return;
			}
			
			int mid = (low + high) / 2;
			int left = node << 1;
			int right = left | 1;
			buildRange(left, low, mid);
			buildRange(right, mid + 1, high);
			tree[node] = max(tree[left], tree[right]);
		}
		
		
		void printSegmentTree() {
			System.out.println(Arrays.toString(tree));
		}
		
	}
	
	private static class KMP{
		
		private static char[] t;
		private static char[] s;
		
		public int kmp(char[]t, char[]s) {
			this.t = t;
			this.s = s;
			return this.kmp();
		}
		
		private int kmp() {
			
			List<Integer> prefixTable = getPrefixTable(s);
			
			int match = 0;
			int i = 0;
			int j = 0;
			
			int n = t.length;
			int m = s.length;
			while(i < n) {
				if(t[i] == s[j]) {
					if(j == m - 1) {
						match++;
						j = prefixTable.get(j - 1);
						continue;
					}
					i++;
					j++;
				}
				else if(j > 0) {
					j = prefixTable.get(j - 1);
				}
				else {
					i++;
				}
			}
			
			return match;
		}

		/***
		1. We compute the prefix values π[i] in a loop by iterating 	<br/>
				from i=1 to i=n−1 (π[0] just gets assigned with 0).		<br/><br/>
		2. To calculate the current value π[i] we set the variable j 	<br/>
				denoting the length of the best suffix for i−1. Initially j=π[i−1].	<br/>
		3. Test if the suffix of length j+1 is also a prefix by		 <br/><br/>
				comparing s[j] and s[i]. If they are equal then we assign π[i]=j+1, 	<br/>
				otherwise we reduce j to π[j−1] and repeat this step.		<br/><br/>
		4. If we have reached the length j=0 and still don't have a match,	<br/> 
				then we assign π[i]=0 and go to the next index i+1.		<br/><br/>
				
		@param
	 	pattern(String)
	***/
		private List<Integer> getPrefixTable(char[] pattern){
			
			List<Integer> prefixTable = new ArrayList<Integer>();
			int n = pattern.length;
			for(int i = 0; i < n; i++) {
				prefixTable.add(0);
			}
			
			for(int i = 1; i < n; i++) {
				for(int j = prefixTable.get(i - 1); j >= 0;) {
					if(pattern[j] == pattern[i]) {
						prefixTable.set(i, j + 1);
						break;
					}
					else if(j > 0){
						j = prefixTable.get(j - 1);
					}
					else {
						break;
					}
				}
			}
			
			return prefixTable;
		}
	}
	


	private static class Point {
		long x;
		long y;
		Point(long x, long y){
			this.x = x;
			this.y = y;
		}
		
		@Override
		public boolean equals(Object obj) {
			Point ob = (Point) obj;
			if(this.x == ob.x && this.y == ob.y){
				return true;
			}
			return false;
			
		}
		@Override
		public String toString() {
			return this.x + " " + this.y;
		}
		
		@Override
		public int hashCode() {
		    return 0;
		}
	}

	private static long pow(int base, int pow) {
		long val = 1L;
		for(int i = 1; i <= pow; i++) {
			val *= base;
		}
		return val;
	}
	
	private static int log(int x, int base) {
		return (int) (Math.log(x) / Math.log(base));
	}
	private static int log(long x, int base) {
		return (int) (Math.log(x) / Math.log(base));
	}
	private static long max(long a, long b) {
		if (a >= b) {
			return a;
		}
		return b;
	}

	private static long abs(long a) {
		if (a < 0) {
			return -a;
		}
		return a;
	}

	private static int abs(int a) {
		if (a < 0) {
			return -a;
		}
		return a;
	}

	private static int max(int a, int b) {
		if (a >= b) {
			return a;
		}
		return b;
	}

	private static int min(int a, int b) {
		if (a <= b) {
			return a;
		}
		return b;
	}
	
	private static long min(long a, long b) {
		if (a <= b) {
			return a;
		}
		return b;
	}

	private static long gcd(long a, long b) {
		if (b == 0) {
			return a;
		}
		return gcd(b, a % b);
	}

	private static int gcd(int a, int b) {
		if (b == 0) {
			return a;
		}
		return gcd(b, a % b);
	}

	private static long bigMod(long n, long k, long m) {

		long ans = 1;
		while (k > 0) {
			if ((k & 1) == 1) {
				ans = (ans * n) % m;
			}
			n = (n * n) % m;
			k >>= 1;
		}
		return ans;
	}

	/*
	 * Returns an iterator pointing to the first element in the range [first,
	 * last] which does not compare less than val.
	 * 
	 */
	private static int lowerBoundNew(long[] arr, long num) {
		int start = 0;
		int end = arr.length - 1;
		int index = 0;
		int len = arr.length;
		int mid = 0;
		while (true) {
			if (start > end) {
				break;
			}
			mid = (start + end) / 2;
			if (arr[mid] > num) {
				end = mid - 1;
			} else if (arr[mid] < num) {
				start = mid + 1;
			} else {
				while (mid >= 0 && arr[mid] == num) {
					mid--;
				}
				return mid + 1;
			}
		}
		if (arr[mid] < num) {
			return mid + 1;
		}
		return mid;
	}

	/*
	 * upper_bound() is a standard library function in C++ defined in the header
	 * . It returns an iterator pointing to the first element in the range
	 * [first, last) that is greater than value, or last if no such element is
	 * found
	 * 
	 */
	private static int upperBoundNew(long arr[], long num) {

		int start = 0;
		int end = arr.length - 1;
		int index = 0;
		int len = arr.length;
		int mid = 0;
		while (true) {
			if (start > end) {
				break;
			}
			mid = (start + end) / 2;
			long val = arr[mid];
			if (val > num) {
				end = mid - 1;
			} else if (val < num) {
				start = mid + 1;
			} else {
				while (mid < len && arr[mid] == num) {
					mid++;
				}
				if (mid == len - 1 && arr[mid] == num) {
					return mid + 1;
				} else {
					return mid;
				}
			}
		}
		if (arr[mid] < num) {
			return mid + 1;
		}
		return mid;
	}

	private static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream));
			tokenizer = null;
		}

		public String next() {
			try {
				while (tokenizer == null || !tokenizer.hasMoreTokens()) {
					tokenizer = new StringTokenizer(reader.readLine());

				}
			} catch (IOException e) {
				return null;
			}
			return tokenizer.nextToken();
		}

		public String nextLine() {
			String line = null;
			try {
				tokenizer = null;
				line = reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
			return line;
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

		public double nextDouble() {
			return Double.parseDouble(next());
		}

		public long nextLong() {
			return Long.parseLong(next());
		}

		public boolean hasNext() {
			try {
				while (tokenizer == null || !tokenizer.hasMoreTokens()) {
					tokenizer = new StringTokenizer(reader.readLine());
				}
			} catch (Exception e) {
				return false;
			}
			return true;
		}
	}
}