从左往右扫描维护上一次遇到 $1$ 的位置，如果当前位置 $i$ 是 $0$，并且距离上一次 $1$ 的位置 $pre$ 的距离大于 $k$ 那么这个点可能是可行的，如果后面的 $1$ 和 $i$ 的距离不大于 $k$，再把 $k$ 点变成不可行的，这样找到的可行的点都是尽量靠左的。

// Date: Tue Dec 12 23:08:09 2023

#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>

#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

using namespace std;

const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
const int dir[8][2] = {
    {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;

const ull Pr = 131;

#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])

#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nonempty(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())

#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z)                                                        \
  cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#else
#define debug1
#define debug2
#define debug3
#endif

const int N = 200010;
int t, n, k;
string s;
bool vis[N];

int main(void) {
#ifdef _DEBUG
  freopen("1367c.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  cin >> t;

  while (t--) {
    cin >> n >> k >> s;
    memset(vis, false, sizeof vis);

    int pre = -1;

    For(i, 0, n) {
      if (s[i] == '1') {
        if (pre != -1 && i - pre <= k) {
          vis[pre] = false;
        }
        pre = i;
      } else {
        if (pre == -1 || i - pre > k) {
          vis[i] = true;
          pre = i;
        }
      }
    }

    int res{};
    For(i, 0, n) {
      if (vis[i])
        res++;
    }

    cout << res << '\n';
  }

  return 0;
}