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William really likes the cellular automaton called "Game of Life" so he decided to make his own version. For simplicity, William decided to define his cellular automaton on an array containing n cells, with each cell either being alive or dead.

Evolution of the array in William's cellular automaton occurs iteratively in the following way:

  * If the element is dead and it has exactly 1 alive neighbor in the current state of the array, then on the next iteration it will become alive. For an element at index i the neighbors would be elements with indices i - 1 and i + 1. If there is no element at that index, it is considered to be a dead neighbor. 
  * William is a humane person so all alive elements stay alive. 



Check the note section for examples of the evolution.

You are given some initial state of all elements and you need to help William find the state of the array after m iterations of evolution.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^3). Description of the test cases follows.

The first line of each test case contains two integers n and m (2 ≤ n ≤ 10^3, 1 ≤ m ≤ 10^9), which are the total number of cells in the array and the number of iterations.

The second line of each test case contains a string of length n made up of characters "0" and "1" and defines the initial state of the array. "1" means a cell is alive and "0" means it is dead.

It is guaranteed that the sum of n over all test cases does not exceed 10^4.

Output

In each test case output a string of length n, made up of characters "0" and "1" — the state of the array after m iterations of evolution.

Example

Input


4
11 3
01000000001
10 2
0110100101
5 2
10101
3 100
000


Output


11111001111
1110111101
10101
000

Note

Sequence of iterations of evolution for the first test case 

  * 01000000001 — initial state 
  * 11100000011 — first iteration of evolution 
  * 11110000111 — second iteration of evolution 
  * 11111001111 — third iteration of evolution 



Sequence of iterations of evolution for the second test case 

  * 0110100101 — initial state 
  * 1110111101 — first iteration of evolution 
  * 1110111101 — second iteration of evolution 
//#pragma optimization_level 3

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <chrono>
#include <ctime>
#include <queue>
#include <math.h>
#include <deque>
#include <stack>
#include <iomanip>
#include <assert.h>
#include <stdio.h>
#include <cstring>
#include <random>

using namespace std;

#define int long long
#define ll long long
#define ull unsigned long long
#define ld long double
#define pii pair<int, int>
#define pld pair<ld, ld>
#define ti tuple<int, int, int>
#define vi vector <int>
#define vpi vector <pii>
#define vld vector <ld>
#define try tr
#define left left11
#define right right11
#define lb lower_bound
#define ub upper_bound
#define sz(c) (int)(c).size()
#define all(c) (c).begin(), (c).end()
#define pb push_back
#define pf push_front
#define sqr(a) ((a) * (a))
#define dbg(x) cout << #x << " : " << x << endl;
#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

#define x first
#define y second

mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());

clock_t START;

inline ld TIME() {
    return (ld) (clock() - START) / CLOCKS_PER_SEC;
}

void SHOW() {
    cout << fixed << setprecision(10);
    cout << TIME() << " SECONDS FROM START\n";
}

ld TL = 2.0;
int calls = 0;

inline bool IS() {
    if (TL - TIME() < 0.1)
        return true;
    return false;
}

template<typename T1, typename T2>
inline bool amin(T1 &a, T2 b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}

template<typename T1, typename T2>
inline bool amax(T1 &a, T2 b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

template<typename T1, typename T2>
ostream &operator<<(ostream &os, pair<T1, T2> &p) {
    os << p.first << ' ' << p.second;
    return os;
}

template<typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) {
    is >> p.first >> p.second;
    return is;
}

template<typename T>
istream &operator>>(istream &is, vector<T> &v) {
    for (auto &u : v) is >> u;
    return is;
}

template<typename T>
ostream &operator<<(ostream &os, vector<T> &v) {
    for (auto &u : v) os << u << ' ';
    return os;
}

int hash_mod[4] = {1000000007, 998244353, 1000000009, 999999937}, mod = hash_mod[rnd() % 4];
int hash_pows[4] = {179, 239, 1007, 2003}, P = hash_pows[rnd() % 4];

int binpow(int x, int p) {
    int res = 1;
    while (p) {
        if (p % 2)
            res = (res * x) % mod;
        x = (x * x) % mod;
        p /= 2;
    }
    return res;
}

const int N = 1e6 + 7, M = 2e3 + 7, lg = 20, sq = 550, inf = 2e9 + 7;
const ld eps = 1e-9, pi = 3.14159265359;

void solve() {
    int n, m;
    cin >> n >> m;
    string a;
    cin >> a;
    string b = a;
    for (int i = 0; i < n; i++) {
        if(a[i] == '1')
            continue;
        int left = m + 1, right = m + 1;
        for (int j = i; j >= 0; j--) {
            if(a[j] == '1') {
                left = i - j;
                break;
            }
        }
        for (int j = i; j < n; j++) {
            if(a[j] == '1') {
                right = j - i;
                break;
            }
        }
        if(left == right)
            continue;
        if(min(left, right) <= m)
            b[i] = '1';
    }
    cout << b << endl;
}

signed main() {
#ifdef _LOCAL
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    fast
    int _;
    cin >> _;
    while(_--)
        solve();
}