Let's call two strings a and b (both of length k) a bit similar if they have the same character in some position, i. e. there exists at least one i ∈ [1, k] such that a_i = b_i.

You are given a binary string s of length n (a string of n characters 0 and/or 1) and an integer k. Let's denote the string s[i..j] as the substring of s starting from the i-th character and ending with the j-th character (that is, s[i..j] = s_i s_{i + 1} s_{i + 2} ... s_{j - 1} s_j).

Let's call a binary string t of length k beautiful if it is a bit similar to all substrings of s having length exactly k; that is, it is a bit similar to s[1..k], s[2..k+1], ..., s[n-k+1..n].

Your goal is to find the lexicographically smallest string t that is beautiful, or report that no such string exists. String x is lexicographically less than string y if either x is a prefix of y (and x ≠ y), or there exists such i (1 ≤ i ≤ min(|x|, |y|)), that x_i < y_i, and for any j (1 ≤ j < i) x_j = y_j.

Input

The first line contains one integer q (1 ≤ q ≤ 10000) — the number of test cases. Each test case consists of two lines.

The first line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 10^6). The second line contains the string s, consisting of n characters (each character is either 0 or 1).

It is guaranteed that the sum of n over all test cases does not exceed 10^6.

Output

For each test case, print the answer as follows:

  * if it is impossible to construct a beautiful string, print one line containing the string NO (note: exactly in upper case, you can't print No, for example); 
  * otherwise, print two lines. The first line should contain the string YES (exactly in upper case as well); the second line — the lexicographically smallest beautiful string, consisting of k characters 0 and/or 1. 

Example

Input


7
4 2
0110
4 2
1001
9 3
010001110
9 3
101110001
10 3
0101110001
10 10
1111111111
11 10
11111111110


Output


YES
11
YES
00
YES
010
YES
101
NO
YES
0000000001
YES
0000000010
import java.io.*;
import java.util.*;

public class PEdu101E {
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(br.readLine());
        StringBuilder sb = new StringBuilder();
        for(int i=0; i<t; i++) {
            String[] line = br.readLine().split(" ");
            int n = Integer.parseInt(line[0]);
            int k = Integer.parseInt(line[1]);
            String s = br.readLine();
            int maxBits = Math.min(getMSB(n), k);
            int max = 1 << maxBits;
            boolean[] notValid = new boolean[max];

            int mask = 1;
            for(int j=1; j<maxBits; j++) {
                mask = mask << 1;
                mask++;
            }

            int[] runningTotal = new int[n+1];
            for(int j=1; j<=n; j++) {
                runningTotal[j] = runningTotal[j-1] + (s.charAt(j-1) == '0' ? 0 : 1);
            }

            int prefixBits = k - maxBits;

            for(int j=Math.max(0, k-maxBits); j<=n-maxBits; j++) {
                if(runningTotal[j] - runningTotal[j-prefixBits] == prefixBits) {
                    String sub = s.substring(j, j+maxBits);
                    int val = ~Integer.parseInt(sub, 2) & mask;
                    if(val < max) {
                        notValid[val] = true;
                    }
                }
            }

            boolean ansFound = false;
            for(int j=0; j<notValid.length; j++) {
                if (!notValid[j]) {
                    sb.append("YES\n");
                    String ans = Integer.toBinaryString(j);
                    for (int x = ans.length(); x < k; x++) {
                        sb.append('0');
                    }
                    sb.append(ans).append("\n");
                    ansFound = true;
                    break;
                }
            }

            if(!ansFound) {
                sb.append("NO\n");
            }
        }
        System.out.print(sb.toString());
        System.out.flush();
    }

    public static int getMSB(int input) {
        int max = 0;
        for(int i=0; i<=20; i++) {
            int mask = 1 << i;
            if((input & mask) == mask) {
                max = i;
            }
        }
        return max + 1;
    }
}
