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.BigDecimal;
import java.math.BigInteger;
import java.util.*;
 
public class Main {
	static final int INF = (int) (1e9 + 10);
	static final int MOD = (int) (1e9 + 7);
 
	// static final int N = (int) (4e5 + 5);
	// static ArrayList<Integer>[] graph;
	// static boolean visited[];
	// static long size[];
 
	public static void main(String[] args) throws NumberFormatException, IOException {
		FastReader sc = new FastReader();
		PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int t = sc.nextInt();
		while (t-- > 0) {
			char[] a = sc.next().toCharArray();
			char[] b = sc.next().toCharArray();
			char[] c = sc.next().toCharArray();
			boolean got = true;
			for(int i =0 ; i < a.length ; i++){
			    if(( b[i] != c[i]) && (a[i] != c[i])){
			        got = false;
			        break;
			    }
			}
			if(got){
			    pr.println("YES");    
			}else{
			    pr.println("NO");
			}
			
 
		}
		pr.flush();
		pr.close();
	}
 
	/*
	 * Template From Here
	 */
	private static boolean possible(long[] arr, int[] f, long mid, long k) {
		long c = 0;
 
		for (int i = 0; i < arr.length; i++) {
			if ((arr[i] * f[f.length - i - 1]) > mid) {
				c += (arr[i] - (mid / f[f.length - i - 1]));
			}
		}
		// System.out.println(mid+" "+c);
		if (c <= k)
			return true;
		return false;
	}
 
	public static int lowerBound(ArrayList<Integer> array, int length, long value) {
		int low = 0;
		int high = length;
		while (low < high) {
			final int mid = (low + high) / 2;
			if (value <= array.get(mid)) {
				high = mid;
			} else {
				low = mid + 1;
			}
		}
		return low;
	}
 
	static class FastReader {
		BufferedReader br;
		StringTokenizer st;
 
		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
 
		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
 
		int nextInt() {
			return Integer.parseInt(next());
		}
 
		long nextLong() {
			return Long.parseLong(next());
		}
 
		double nextDouble() {
			return Double.parseDouble(next());
		}
 
		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}
 
	static long gcd(long a, long b) {
		if (a == 0)
			return b;
		return gcd(b % a, a);
	}
 
	static long lcm(long a, long b) {
		return (a * b) / gcd(a, b);
	}
 
	public static void sortbyColumn(int[][] att, int col) {
		// Using built-in sort function Arrays.sort
		Arrays.sort(att, new Comparator<int[]>() {
 
			@Override
			// Compare values according to columns
			public int compare(final int[] entry1, final int[] entry2) {
 
				// To sort in descending order revert
				// the '>' Operator
				// if (entry1[col] > entry2[col])
				// return 1;
				// else //(entry1[col] >= entry2[col])
				// return -1;
				return new Integer(entry1[col]).compareTo(entry2[col]);
				// return entry1[col].
			}
		}); // End of function call sort().
	}
 
	static class pair {
		int V, I;
 
		pair(int v, int i) {
			V = v;
			I = i;
		}
	}
 
	public static int[] swap(int data[], int left, int right) {
		int temp = data[left];
		data[left] = data[right];
		data[right] = temp;
		return data;
	}
 
	public static int[] reverse(int data[], int left, int right) {
		while (left < right) {
			int temp = data[left];
			data[left++] = data[right];
			data[right--] = temp;
		}
		return data;
	}
 
	public static boolean findNextPermutation(int data[]) {
		if (data.length <= 1)
			return false;
 
		int last = data.length - 2;
		while (last >= 0) {
			if (data[last] < data[last + 1]) {
				break;
			}
			last--;
		}
		if (last < 0)
			return false;
 
		int nextGreater = data.length - 1;
		for (int i = data.length - 1; i > last; i--) {
			if (data[i] > data[last]) {
				nextGreater = i;
				break;
			}
		}
		data = swap(data, nextGreater, last);
		data = reverse(data, last + 1, data.length - 1);
		return true;
	}
 
	static long nCr(long n, long r) {
		if (n == r)
			return 1;
		return fact(n) / (fact(r) * fact(n - r));
	}
 
	static long fact(long n) {
		long res = 1;
		for (long i = 2; i <= n; i++)
			res = res * i;
		return res;
	}
 
	/*
	 * static boolean prime[] = new boolean[1000007];
	 * 
	 * public static void sieveOfEratosthenes(int n) {
	 * 
	 * for (int i = 0; i < n; i++) prime[i] = true; * for (int p = 2; p * p <=
	 * n; p++) { // If prime[p] is not changed, then it is a prime if (prime[p]
	 * == true) { // Update all multiples of p for (int i = p * p; i <= n; i +=
	 * p) prime[i] = false; } }
	 * 
	 * // Print all prime numbers // for(int i = 2; i <= n; i++) // { //
	 * if(prime[i] == true) // System.out.print(i + " "); // } }
	 */
 
	static long power(long fac2, int y, int p) {
		long res = 1;
		fac2 = fac2 % p;
		while (y > 0) {
			if (y % 2 == 1)
				res = (res * fac2) % p;
			fac2 = (fac2 * fac2) % p;
		}
		return res;
	}
 
	static long modInverse(long fac2, int p) {
		return power(fac2, p - 2, p);
	}
 
	static boolean isPrime(int n) {
		// Corner cases
		if (n <= 1)
			return false;
		if (n <= 3)
			return true;
 
		// This is checked so that we can skip
		// middle five numbers in below loop
		if (n % 2 == 0 || n % 3 == 0)
			return false;
 
		for (int i = 5; i * i <= n; i = i + 6)
			if (n % i == 0 || n % (i + 2) == 0)
				return false;
 
		return true;
	}
 
	public static BigInteger lcmm(String a, String b) {
		BigInteger s = new BigInteger(a);
		BigInteger s1 = new BigInteger(b);
		BigInteger mul = s.multiply(s1);
		BigInteger gcd = s.gcd(s1);
		BigInteger lcm = mul.divide(gcd);
		return lcm;
	}
 
}