At the school where Vasya is studying, preparations are underway for the graduation ceremony. One of the planned performances is a ball, which will be attended by pairs of boys and girls.

Each class must present two couples to the ball. In Vasya's class, a boys and b girls wish to participate. But not all boys and not all girls are ready to dance in pairs.

Formally, you know k possible one-boy-one-girl pairs. You need to choose two of these pairs so that no person is in more than one pair.

For example, if a=3, b=4, k=4 and the couples (1, 2), (1, 3), (2, 2), (3, 4) are ready to dance together (in each pair, the boy's number comes first, then the girl's number), then the following combinations of two pairs are possible (not all possible options are listed below): 

  * (1, 3) and (2, 2); 
  * (3, 4) and (1, 3); 



But the following combinations are not possible: 

  * (1, 3) and (1, 2) — the first boy enters two pairs; 
  * (1, 2) and (2, 2) — the second girl enters two pairs; 



Find the number of ways to select two pairs that match the condition above. Two ways are considered different if they consist of different pairs.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

The first line of each test case contains three integers a, b and k (1 ≤ a, b, k ≤ 2 ⋅ 10^5) — the number of boys and girls in the class and the number of couples ready to dance together.

The second line of each test case contains k integers a_1, a_2, … a_k. (1 ≤ a_i ≤ a), where a_i is the number of the boy in the pair with the number i.

The third line of each test case contains k integers b_1, b_2, … b_k. (1 ≤ b_i ≤ b), where b_i is the number of the girl in the pair with the number i.

It is guaranteed that the sums of a, b, and k over all test cases do not exceed 2 ⋅ 10^5.

It is guaranteed that each pair is specified at most once in one test case.

Output

For each test case, on a separate line print one integer — the number of ways to choose two pairs that match the condition above.

Example

Input


3
3 4 4
1 1 2 3
2 3 2 4
1 1 1
1
1
2 2 4
1 1 2 2
1 2 1 2


Output


4
0
2

Note

In the first test case, the following combinations of pairs fit: 

  * (1, 2) and (3, 4); 
  * (1, 3) and (2, 2); 
  * (1, 3) and (3, 4); 
  * (2, 2) and (3, 4). 



There is only one pair in the second test case.

In the third test case, the following combinations of pairs fit: 

  * (1, 1) and (2, 2); 
  * (1, 2) and (2, 1). 
/*
								"जय श्री कृष्णा"
*/
import java.util.*;
import java.io.*;

public class Balls_in_berland_c implements Runnable {

	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;
		}

		int[] nextIntArray(int size) {
			int[] arr = new int[size];
			for (int i = 0; i < size; ++i)
				arr[i] = nextInt();
			return arr;
		}

		long[] nextLongArray(int size) {
			long[] arr = new long[size];
			for (int i = 0; i < size; ++i)
				arr[i] = nextLong();
			return arr;
		}

		double[] nextDoubleArray(int size) {
			double[] arr = new double[size];
			for (int i = 0; i < size; ++i)
				arr[i] = nextDouble();
			return arr;
		}
	}

	static int mod = (int) (1e9 + 7);
	static final PrintWriter out = new PrintWriter(System.out);
	static final FastReader fr = new FastReader();

	public static void main(String[] args) throws java.lang.Exception {
		new Thread(null, new Balls_in_berland_c(), "Main", 1 << 26).start();
	}

	void solve() {
		int a = fr.nextInt(),b=fr.nextInt(),k=fr.nextInt();
		int bo[] = fr.nextIntArray(k),gi[]=fr.nextIntArray(k);
		if(k<2) {
			out.print(0);
			return;
		}
		int pb[]=new int[k],pg[]=new int[k];
		int fb[]=new int[a+1],fg[]=new int[b+1];
		for(int i=k-1; i>=0; i--) {
			pb[i]=++fb[bo[i]]-1;
			pg[i]=++fg[gi[i]]-1;
		}
		long ans=0;
		for(int i=0; i<k;  i++) {
			ans+=(k-(i+1)-(pb[i]+pg[i]));
		}
		out.print(ans);
	}

	@Override
	public void run() {
		long start = System.nanoTime(); // Program Start
		boolean testcase = true;
		int t = testcase ? fr.nextInt() : 1;
		while (t-- > 0) {
			solve();
			out.print("\n");
			out.flush();
		}
		long end = System.nanoTime(); // Program End
		System.err.println("Time taken: " + (end - start) / 1000000 + " ms");
	}

	static class CP {

		static boolean isPrime(long n) {
			if (n <= 1)
				return false;
			if (n == 2 || n == 3)
				return true;
			if (n % 2 == 0 || n % 3 == 0)
				return false;
			for (int i = 5; i * i <= n; i += 6) {
				if (n % i == 0 || n % (i + 2) == 0)
					return false;
			}
			return true;
		}

		static long binary_Expo(long a, long b) { // calculating a^b
			long res = 1;
			while (b != 0) {
				if ((b & 1) == 1) {
					res *= a;
					--b;
				}
				a *= a;
				b /= 2;
			}
			return res;
		}

		static long Modular_Expo(long a, long b) {
			long res = 1;
			while (b != 0) {
				if ((b & 1) == 1) {
					res = (res * a) % mod;
					--b;
				}
				a = (a * a) % mod;
				b /= 2;
			}
			return res;
		}

		static int i_gcd(int a, int b) {// iterative way to calculate gcd.
			while (true) {
				if (b == 0)
					return a;
				int c = a;
				a = b;
				b = c % b;
			}
		}

		static long gcd(long a, long b) {// here b is the remainder
			if (b == 0)
				return a; //because each time b will divide a.
			return gcd(b, a % b);
		}

		static long ceil_div(long a, long b) { // a numerator b denominator
			return (a + b - 1) / b;
		}

		static int getIthBitFromInt(int bits, int i) {
			return (bits >> (i - 1)) & 1;
		}

		static int upper_Bound(int a[], int x) {//closest to the left+1
			int l = -1, r = a.length;
			while (l + 1 < r) {
				int m = (l + r) >>> 1;
				if (a[m] <= x)
					l = m;
				else
					r = m;
			}
			return l + 1;
		}

		static int lower_Bound(int a[], int x) {//closest to the right
			int l = -1, r = a.length;
			while (l + 1 < r) {
				int m = (l + r) >>> 1;
				if (a[m] >= x)
					r = m;
				else
					l = m;
			}
			return r;
		}

		static void sort(int a[]) {
			PriorityQueue<Integer> q = new PriorityQueue<>();
			for (int i = 0; i < a.length; i++)
				q.add(a[i]);
			for (int i = 0; i < a.length; i++)
				a[i] = q.poll();
		}

		static int getMax(int arr[], int n) {
			int mx = arr[0];
			for (int i = 1; i < n; i++)
				if (arr[i] > mx)
					mx = arr[i];
			return mx;
		}
	}

}
