Problem: You are given string s of length n consisting of 0-s and 1-s. You build an infinite string t as a concatenation of an infinite number of strings s, or t = ssss ... For example, if s = 10010, then t = 100101001010010...

Calculate the number of prefixes of t with balance equal to x. The balance of some string q is equal to cnt_{0, q} - cnt_{1, q}, where cnt_{0, q} is the number of occurrences of 0 in q, and cnt_{1, q} is the number of occurrences of 1 in q. The number of such prefixes can be infinite; if it is so, you must say that.

A prefix is a string consisting of several first letters of a given string, without any reorders. An empty prefix is also a valid prefix. For example, the string "abcd" has 5 prefixes: empty string, "a", "ab", "abc" and "abcd".

Input

The first line contains the single integer T (1 ≤ T ≤ 100) — the number of test cases.

Next 2T lines contain descriptions of test cases — two lines per test case. The first line contains two integers n and x (1 ≤ n ≤ 10^5, -10^9 ≤ x ≤ 10^9) — the length of string s and the desired balance, respectively.

The second line contains the binary string s (|s| = n, s_i ∈ \{0, 1\}).

It's guaranteed that the total sum of n doesn't exceed 10^5.

Output

Print T integers — one per test case. For each test case print the number of prefixes or -1 if there is an infinite number of such prefixes.

Example

Input


4
6 10
010010
5 3
10101
1 0
0
2 0
01


Output


3
0
1
-1

Note

In the first test case, there are 3 good prefixes of t: with length 28, 30 and 32.

Solution (in JAVA):

Solution:
import java.util.*;

public class Q2 {

    static void calMax(int[] diff, int[] max) {
        int maxPos = 1;
        int maxNeg = -1;
        for (int i = 0; i < diff.length; i++) {
            if (diff[i] > maxPos) {
                maxPos = diff[i];
            } else if (diff[i] < maxNeg) {
                maxNeg = diff[i];
            }
        }
        max[0] = maxPos;
        max[1] = maxNeg;
    }

    static int count(String s, int balance, int lastVal, Map<Integer, Integer> diff) {
        int count = 0;
        for (int i = 0; i < s.length(); i++) {
            int d = balance - diff.get(i);
            if (d / lastVal >= 0) {
                    count++;
            }
        }
        return count;
    }

    static int check(int len, int balance, String s) {
        int count = 0;

        Map<Integer, Integer> diff = new HashMap<>();
        Set<Integer> values = new HashSet<>();

        if (s.charAt(0) == '1') {
            diff.put(0, -1);
            values.add(-1);
        } else {
            diff.put(0, 1);
            values.add(1);
        }
        for (int i = 1; i < len; i++) {
            int tmp = diff.get(i - 1);
            if (s.charAt(i) == '0') {
                diff.put(i, tmp + 1);
                values.add(tmp + 1);
            } else {
                diff.put(i, diff.get(i - 1) - 1);
                values.add(tmp - 1);
            }
        }

        // calculate the balance of S
        int lastVal = diff.get(diff.size() - 1);

        if (lastVal == 0) {                     // cause infinite prefixes
            if (values.contains(balance)) {
                return -1;
            } else {
                return 0;
            }
        } else if (balance == 0) {
            count++;      // empty string
            if (!values.contains(0)) {
                return 1;       // empty string
            }
        }

        for (int i = 0; i < s.length(); i++) {
            int d = balance - diff.get(i);
            if (d % lastVal == 0) {
                if (d / lastVal >= 0) {
                    count++;
                }
            }
        }

        return count;
    }

    public static void main(String[] arg) {
//   System.out.println("t0: " + check(1, 0, "0"));                // 1
//        System.out.println("t1: " + check(6, 10, "010010"));     // 3
//
//        System.out.println("t2: " + check(5, 3, "10101"));       // 0
//        System.out.println("t3: " + check(2, 0, "01"));          // -1
//        System.out.println("t4: " + check(1, -1100101, "1"));    // 1
//
//        System.out.println("t5: " + check(1, -1100101, "0"));    // 0
//
//        System.out.println("t6: " + check(6, 353709959, "110000")); // 3
//
//        check(5, 0, "10010");
//        System.out.println("t7: " + check(3, 0, "011")); // 3
//          System.out.println("t8: " + check(5, 0, "10010")); // 4
//          System.out.println("t8: " + check(2223, 0, "00100111000000111010111101101001111010101010100101011100100011010111110010010000010000001101110010001000011010110101110011101011010001000111010000111101101100100010010101001000101110010000011100010001011101111101110101100101100111110001001000110101110110010000000111010011100100001100110000101110011010101100101101001101110000000100100001110110101110011000111011011100010100100101100101100010111100000001001010000100011100011100100000000111010000011000100001011100111011000110100011000101011010101000")); // 3



        Scanner sc = new Scanner(System.in);
        Integer testcases = sc.nextInt();
        sc.nextLine();
        int n, x;
        String s;
        for (int i = 0; i < testcases; i++) {
            n = sc.nextInt();
            x = sc.nextInt();
            s = sc.next();
            int res = check(n, x, s);
            System.out.println(res);
            sc.nextLine();
        }
    }
}