Phoenix has a string s consisting of lowercase Latin letters. He wants to distribute all the letters of his string into k non-empty strings a_1, a_2, ..., a_k such that every letter of s goes to exactly one of the strings a_i. The strings a_i do not need to be substrings of s. Phoenix can distribute letters of s and rearrange the letters within each string a_i however he wants.

For example, if s =  baba and k=2, Phoenix may distribute the letters of his string in many ways, such as: 

  * ba and ba
  * a and abb
  * ab and ab
  * aa and bb



But these ways are invalid: 

  * baa and ba
  * b and ba
  * baba and empty string (a_i should be non-empty) 



Phoenix wants to distribute the letters of his string s into k strings a_1, a_2, ..., a_k to minimize the lexicographically maximum string among them, i. e. minimize max(a_1, a_2, ..., a_k). Help him find the optimal distribution and print the minimal possible value of max(a_1, a_2, ..., a_k).

String x is lexicographically less than string y if either x is a prefix of y and x ≠ y, or there exists an index i (1 ≤ i ≤ min(|x|, |y|)) such that x_i < y_i and for every j (1 ≤ j < i) x_j = y_j. Here |x| denotes the length of the string x.

Input

The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Each test case consists of two lines.

The first line of each test case consists of two integers n and k (1 ≤ k ≤ n ≤ 10^5) — the length of string s and the number of non-empty strings, into which Phoenix wants to distribute letters of s, respectively.

The second line of each test case contains a string s of length n consisting only of lowercase Latin letters.

It is guaranteed that the sum of n over all test cases is ≤ 10^5.

Output

Print t answers — one per test case. The i-th answer should be the minimal possible value of max(a_1, a_2, ..., a_k) in the i-th test case.

Example

Input


6
4 2
baba
5 2
baacb
5 3
baacb
5 3
aaaaa
6 4
aaxxzz
7 1
phoenix


Output


ab
abbc
b
aa
x
ehinopx

Note

In the first test case, one optimal solution is to distribute baba into ab and ab. 

In the second test case, one optimal solution is to distribute baacb into abbc and a.

In the third test case, one optimal solution is to distribute baacb into ac, ab, and b.

In the fourth test case, one optimal solution is to distribute aaaaa into aa, aa, and a.

In the fifth test case, one optimal solution is to distribute aaxxzz into az, az, x, and x.

In the sixth test case, one optimal solution is to distribute phoenix into ehinopx.
// package name;
import java.io.*;
import java.util.*;

public class Main {
	static final long mod = 1000000007;
	static final long lmx = 9223372036854775807l, lmn = -9223372036854775808l;
	static final int imx = 2147483647, imn = -2147483648;
	static final String Y = "YES", NO = "NO", LN = "\n", SPC = " ";
	static Scanner s = new Scanner(System.in);
	//static Reader s = new Reader();
	static PrintWriter out = new PrintWriter(System.out);

	public static void main(String[] args) throws IOException {
		int t = 1;
		//System.out.println(-1337%3);
		t = s.nextInt();
		String xy = slinp();
		StringBuffer sb = new StringBuffer();
		while (t > 0) {
			t--;
			int n = iinp(), k = iinp();
			xy = slinp();
			String str = slinp();
			int a[] = new int[26];
			for(int i = 0; i < str.length(); i++) {
				a[str.charAt(i) - 'a']++;
			}
			int sm = 0, co = 0;
			for(int i : a) if (i != 0) co++;
			for(int i = 0; i < 26; i++) 
				if(a[i] != 0) {
					sm = i;
					break;
				}
			if(a[sm] < k) {
				char kthSCh = (char)('a' + sm);
				int c = a[sm];
				for(int i = sm + 1; i < 26; i++) {
					if(a[i] != 0) {
						kthSCh = (char)('a' + i);
						c += a[i];
					}
					if(c >= k) {
						break;
					}
				}
				sb.append(kthSCh + LN);
			}
			else {
				if(co == 1) {
					char ch = (char) ('a' + sm);
					int val = a[sm] / k;
					if(a[sm] % k > 0) val++;
					while(val > 0) {
						sb.append(ch);
						val--;
					}
				}
				else if(co == 2 && a[sm] == k) {
					sb.append((char)('a' + sm));
					for(int i = sm + 1; i < 26; i++)
						if(a[i] > 0) {
							char ch = (char) ('a' + i);
							int val = a[i] / k;
							if(a[i] % k > 0) val++;
							while(val > 0) {
								sb.append(ch);
								val--;
							}
						}
				}
				else {
					a[sm] -= k - 1;
					for(int i = sm; i < 26; i++)
						while(a[i] > 0) {
							a[i]--;
							sb.append((char)('a' + i));
						}
				}
				sb.append(LN);
			}
		}
		out.print(sb);
		out.flush();
	}

	static int iinp() throws IOException {
		return s.nextInt();
	}

	static long linp() throws IOException {
		return s.nextLong();
	}

	static double dinp() throws IOException {
		return s.nextDouble();
	}

	static String sinp() throws IOException {
		return s.next();
	}

	static String slinp() throws IOException {
		return s.nextLine();
	}

	static int[] iarr(int n) {
		return new int[n];
	}

	static long[] larr(int n) {
		return new long[n];
	}

	static double[] darr(int n) {
		return new double[n];
	}

	static boolean[] barr(int n) {
		return new boolean[n];
	}

	static String[] sarr(int n) {
		return new String[n];
	}

	static int[][] iarr(int n, int m) {
		return new int[n][m];
	}

	static long[][] larr(int n, int m) {
		return new long[n][m];
	}

	static int[] iinp(int n, int st, int size) throws IOException {
		int a[] = iarr(n);
		for (int i = st; i < st + size; i++)
			a[i] = iinp();
		return a;
	}

	static long[] linp(int n, int st, int size) throws IOException {
		long a[] = larr(n);
		for (int i = st; i < st + size; i++)
			a[i] = linp();
		return a;
	}

	static String[] sinp(int n, int st, int size, String str) throws IOException {
		String a[] = sarr(n);
		for (int i = st; i < st + size; i++)
			if (str.equalsIgnoreCase(SPC))
				a[i] = sinp();
			else
				a[i] = slinp();
		return a;
	}

	static int[][] iinp(int n, int m, int row, int rows, int col, int cols) throws IOException {
		int a[][] = iarr(n, m);
		for (int i = row; i < row + rows; i++)
			for (int j = col; j < col + cols; j++)
				a[i][j] = iinp();
		return a;
	}

	static long[][] linp(int n, int m, int row, int rows, int col, int cols) throws IOException {
		long a[][] = larr(n, m);
		for (int i = row; i < row + rows; i++)
			for (int j = col; j < col + cols; j++)
				a[i][j] = linp();
		return a;
	}

	static void ginp(ArrayList<Integer> g[], int m) throws IOException {
		for (int i = 0; i < m; i++) {
			int u = iinp(), v = iinp();
			g[u].add(u);
			g[v].add(u);
		}
	}

	static long min(long... a) {
		long min = lmx;//Long.MAX_VALUE;
		for (long i : a)
			if (i < min)
				min = i;
		return min;
	}

	static int min(int... a) {
		int min = imx;//Long.MAX_VALUE;
		for (int i : a)
			if (i < min)
				min = i;
		return min;
	}

	static long max(long... a) {
		long max = lmn;//Long.MIN_VALUE;
		for (long i : a)
			if (i > max)
				max = i;
		return max;
	}

	static int max(int... a) {
		int max = imn;//Long.MIN_VALUE;
		for (int i : a)
			if (i > max)
				max = i;
		return max;
	}

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

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

	static void sort(int a[]) {
		Arrays.sort(a);
	};

	static void sort(long a[]) {
		Arrays.sort(a);
	};

	static void sort(String a[]) {
		Arrays.sort(a);
	};

	static void sort(int a[], int st, int e) {
		Arrays.sort(a, st, e);
	};

	static void sort(long a[], int st, int e) {
		Arrays.sort(a, st, e);
	};

	static void sort(String a[], int st, int e) {
		Arrays.sort(a, st, e);
	};

	static void sortbycol(int arr[][]) {
		Arrays.sort(arr, new Comparator<int[]>() {
			public int compare(int[] x, int[] y) {
				if (x[1] == y[1])
					return x[2] - y[2];
				return x[1] - y[1]; // sort in increasing order on the basis of first column
			}
		});
	}

	static StringBuffer printarr(long a[], String str) {
		StringBuffer ans = new StringBuffer();
		for (long i : a)
			ans.append(i + str);
		return ans;
	}

}

class Pair implements Comparator<Pair> {
	long a, b;

	Pair(long x, long y) {
		a = x;
		b = y;
	}

	public int compare(Pair p1, Pair p2) {
		if (p1.a != p2.a)
			return (p1.a - p2.a > 0) ? 1 : -1;
		return (p1.b - p2.b > 0) ? 1 : -1;
	}
}
