Problem: AquaMoon had n strings of length m each. n is an odd number.

When AquaMoon was gone, Cirno tried to pair these n strings together. After making (n-1)/(2) pairs, she found out that there was exactly one string without the pair!

In her rage, she disrupted each pair of strings. For each pair, she selected some positions (at least 1 and at most m) and swapped the letters in the two strings of this pair at the selected positions.

For example, if m = 6 and two strings "abcdef" and "xyzklm" are in one pair and Cirno selected positions 2, 3 and 6 she will swap 'b' with 'y', 'c' with 'z' and 'f' with 'm'. The resulting strings will be "ayzdem" and "xbcklf".

Cirno then stole away the string without pair and shuffled all remaining strings in arbitrary order.

AquaMoon found the remaining n-1 strings in complete disarray. Also, she remembers the initial n strings. She wants to know which string was stolen, but she is not good at programming. Can you help her?

Input

This problem is made as interactive. It means, that your solution will read the input, given by the interactor. But the interactor will give you the full input at the beginning and after that, you should print the answer. So you should solve the problem, like as you solve the usual, non-interactive problem because you won't have any interaction process. The only thing you should not forget is to flush the output buffer, after printing the answer. Otherwise, you can get an "Idleness limit exceeded" verdict. Refer to the [interactive problems guide](https://codeforces.com/blog/entry/45307) for the detailed information about flushing the output buffer.

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

The first line of each test case contains two integers n, m (1 ≤ n ≤ 10^5, 1 ≤ m ≤ 10^5) — the number of strings and the length of each string, respectively.

The next n lines each contain a string with length m, describing the original n strings. All string consists of lowercase Latin letters.

The next n-1 lines each contain a string with length m, describing the strings after Cirno exchanged and reordered them.

It is guaranteed that n is odd and that the sum of n ⋅ m over all test cases does not exceed 10^5.

Hack format:

The first line should contain a single integer t. After that t test cases should follow in the following format:

The first line should contain two integers n and m.

The following n lines should contain n strings of length m, describing the original strings.

The following (n-1)/(2) lines should describe the pairs. They should contain, in the following order: the index of the first string i (1 ≤ i ≤ n), the index of the second string j (1 ≤ j ≤ n, i ≠ j), the number of exchanged positions k (1 ≤ k ≤ m), and the list of k positions that are exchanged (k distinct indices from 1 to m in any order).

The final line should contain a permutation of integers from 1 to n, describing the way the strings should be reordered. The strings will be placed in the order indices placed in this permutation, the stolen string index will be ignored.

Output

For each test case print a single line with the stolen string.

Example

Input


3
3 5
aaaaa
bbbbb
ccccc
aaaaa
bbbbb
3 4
aaaa
bbbb
cccc
aabb
bbaa
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
xbcklf
eueueu
ayzdem
ukukuk


Output


ccccc
cccc
kekeke

Note

In the first test case, "aaaaa" and "bbbbb" exchanged all positions, and "ccccc" is the stolen string.

In the second test case, "aaaa" and "bbbb" exchanged two first positions, and "cccc" is the stolen string.

This is the first test in the hack format: 
    
    
      
    3  
    3 5  
    aaaaa  
    bbbbb  
    ccccc  
    1 2 5 1 2 3 4 5  
    2 1 3  
    3 4  
    aaaa  
    bbbb  
    cccc  
    1 2 2 1 2  
    2 1 3  
    5 6  
    abcdef  
    uuuuuu  
    kekeke  
    ekekek  
    xyzklm  
    1 5 3 2 3 6  
    2 4 3 2 4 6  
    5 4 1 2 3  
    

Solution (in JAVA):

Solution:
//Utilities
import java.io.*;
import java.util.*;

public class B {
	static int t;
	static int n, m;
	static char[][] a, b;
	static char[] ch;
	static int[][] cntA, cntB, cntDif; // [pos][letter]
	static char[] letterNeed;
	
	public static void main(String[] args) throws IOException {
		t = in.iscan();
		outer : while(t-- > 0) {
			n = in.iscan(); m = in.iscan();
			a = new char[n][m]; b = new char[n-1][m];
			cntA = new int[m][26]; cntB = new int[m][26]; cntDif = new int[m][26];
			for (int i = 0; i < n; i++) {
				a[i] = in.sscan().toCharArray();
				for (int j = 0; j < m; j++) {
					cntA[j][a[i][j]-'a']++;
				}
			}
			for (int i = 0; i < n-1; i++) {
				b[i] = in.sscan().toCharArray();
				for (int j = 0; j < m; j++) {
					cntB[j][b[i][j]-'a']++;
				}
			}
			letterNeed = new char[m];
			for (int pos = 0; pos < m; pos++) {
				for (int i = 0; i < 26; i++) {
					cntDif[pos][i] = cntA[pos][i] - cntB[pos][i];
					if (cntDif[pos][i] == 1) {
						letterNeed[pos] = (char)(i+'a');
					}
				}
			}
			for (int i = 0; i < n; i++) {
				boolean flag = true;
				for (int j = 0; j < m; j++) {
					if (a[i][j] != letterNeed[j]) {
						flag = false;
						break;
					}
				}
				if (flag) {
					out.println(a[i]);
					out.flush();
					continue outer;
				}
			}
			out.println(cntA[-1][-1]);
		}
		out.close();
	} 
	
	static INPUT in = new INPUT(System.in);
	static PrintWriter out = new PrintWriter(System.out);
	private static class INPUT {

		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar, numChars;

		public INPUT (InputStream stream) {
			this.stream = stream;
		}

		public INPUT (String file) throws IOException {
			this.stream = new FileInputStream (file);
		}

		public int cscan () throws IOException {
			if (curChar >= numChars) {
				curChar = 0;
				numChars = stream.read (buf);
			}
			
			if (numChars == -1)
				return numChars;

			return buf[curChar++];
		}

		public int iscan () throws IOException {
			int c = cscan (), sgn = 1;
			
			while (space (c))
				c = cscan ();

			if (c == '-') {
				sgn = -1;
				c = cscan ();
			}

			int res = 0;

			do {
				res = (res << 1) + (res << 3);
				res += c - '0';
				c = cscan ();
			}
			while (!space (c));

			return res * sgn;
		}

		public String sscan () throws IOException {
			int c = cscan ();
			
			while (space (c))
				c = cscan ();

			StringBuilder res = new StringBuilder ();

			do {
				res.appendCodePoint (c);
				c = cscan ();
			}
			while (!space (c));

			return res.toString ();
		}

		public double dscan () throws IOException {
			int c = cscan (), sgn = 1;
			
			while (space (c))
				c = cscan ();

			if (c == '-') {
				sgn = -1;
				c = cscan ();
			}

			double res = 0;

			while (!space (c) && c != '.') {
				if (c == 'e' || c == 'E')
					return res * UTILITIES.fast_pow (10, iscan ());
				
				res *= 10;
				res += c - '0';
				c = cscan ();
			}

			if (c == '.') {
				c = cscan ();
				double m = 1;

				while (!space (c)) {
					if (c == 'e' || c == 'E')
						return res * UTILITIES.fast_pow (10, iscan ());

					m /= 10;
					res += (c - '0') * m;
					c = cscan ();
				}
			}

			return res * sgn;
		}

		public long lscan () throws IOException {
			int c = cscan (), sgn = 1;
			
			while (space (c))
				c = cscan ();

			if (c == '-') {
				sgn = -1;
				c = cscan ();
			}

			long res = 0;

			do {
				res = (res << 1) + (res << 3);
				res += c - '0';
				c = cscan ();
			}
			while (!space (c));

			return res * sgn;
		}

		public boolean space (int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
	}

	public static class UTILITIES {

		static final double EPS = 10e-6;

		public static int lower_bound (int[] arr, int x) {
			int low = 0, high = arr.length, mid = -1;

			while (low < high) {
				mid = (low + high) / 2;

				if (arr[mid] >= x)
					high = mid;
				else
					low = mid + 1;
			}

			return low;
		}

		public static int upper_bound (int[] arr, int x) {
			int low = 0, high = arr.length, mid = -1;

			while (low < high) {
				mid = (low + high) / 2;

				if (arr[mid] > x)
					high = mid;
				else
					low = mid + 1;
			}

			return low;
		}

		public static int gcd (int a, int b) {
			return b == 0 ? a : gcd (b, a % b);
		}

		public static int lcm (int a, int b) {
			return a * b / gcd (a, b);
		}

		public static long fast_pow_mod (long b, long x, int mod) {
			if (x == 0) return 1;
			if (x == 1) return b;
			if (x % 2 == 0) return fast_pow_mod (b * b % mod, x / 2, mod) % mod;

			return b * fast_pow_mod (b * b % mod, x / 2, mod) % mod;
		}

		public static int fast_pow (int b, int x) {
			if (x == 0) return 1;
			if (x == 1) return b;
			if (x % 2 == 0) return fast_pow (b * b, x / 2);

			return b * fast_pow (b * b, x / 2);
		}

		public static long choose (long n, long k) {
			k = Math.min (k, n - k);
			long val = 1;

			for (int i = 0; i < k; ++i)
				val = val * (n - i) / (i + 1);

			return val;
		}

		public static long permute (int n, int k) {
			if (n < k) return 0;
			long val = 1;

			for (int i = 0; i < k; ++i)
				val = (val * (n - i));

			return val;
		}
		
		// start of permutation and lower/upper bound template
		public static void nextPermutation(int[] nums) {
		    //find first decreasing digit
		    int mark = -1;
		    for (int i = nums.length - 1; i > 0; i--) {
		        if (nums[i] > nums[i - 1]) {
		            mark = i - 1;
		            break;
		        }
		    }
		 
		    if (mark == -1) {
		        reverse(nums, 0, nums.length - 1);
		        return;
		    }
		 
		    int idx = nums.length-1;
		    for (int i = nums.length-1; i >= mark+1; i--) {
		        if (nums[i] > nums[mark]) {
		            idx = i;
		            break;
		        }
		    }
		 
		    swap(nums, mark, idx);
		 
		    reverse(nums, mark + 1, nums.length - 1);
		}
		 
		public static void swap(int[] nums, int i, int j) {
		    int t = nums[i];
		    nums[i] = nums[j];
		    nums[j] = t;
		}
		 
		public static void reverse(int[] nums, int i, int j) {
		    while (i < j) {
		        swap(nums, i, j);
		        i++;
		        j--;
		    }
		}
		
		static int lower_bound (int[] arr, int hi, int cmp) {
			int low = 0, high = hi, mid = -1;
			while (low < high) {
				mid = (low + high) / 2;
				if (arr[mid] >= cmp) high = mid;
				else low = mid + 1;
			}
			return low;
		}
	 
		static int upper_bound (int[] arr, int hi, int cmp) {
			int low = 0, high = hi, mid = -1;
			while (low < high) {
				mid = (low + high) / 2;
				if (arr[mid] > cmp) high = mid;
				else low = mid + 1;
			}
			return low;
		}
		// end of permutation and lower/upper bound template
	}
}