#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, int>;
const int N = 2e6 + 5;
const ll P = 1e9 + 7;
ll m, n, q, k;
// dp[i][j][2]前i个字符的4有j个mid的需要的最小值,1表示连续的
/// 0 dp[i][j][0]=dp[i][j]
string s;
bool ck(int mid) {
    vector<vector<vector<int>>> dp(
        n + 2, vector<vector<int>>(k + 2, vector<int>(2, 1e9)));
    dp[0][0][0] = 0;
    vector<int> pre(n + 2);
    for (int i = 1; i <= n; i++) {
        pre[i] += pre[i - 1];
        if (s[i] == '1') {
            pre[i]++;
        }
        for (int j = 0; j <= k; j++) {
            dp[i][j][0] = min(dp[i - 1][j][0], dp[i - 1][j][1]);
           // dp[i][j][1] = min(dp[i - 1][j][1], dp[i - 1][j][0]);
            if (i >= mid && j >= 1) {
                dp[i][j][1] =
                    min(dp[i - mid][j - 1][0] + pre[i] - pre[mid], dp[i][j][1]);
            }
        }
    }
    if (dp[n][k][0] <= m || dp[n][k][1] <= m) {
        return 1;
    }
    return 0;
}
void solve() {
    cin >> n >> m >> k;
    int l = 1, r = n;
    cin >> s;
    s = ' ' + s;
    while (l <= r) {
        int mid = l + r >> 1;
        if (ck(mid)) {
            l = mid + 1;
        } else {
            r = mid - 1;
        }
    }
    cout << (l - 1 == 0 ? -1 : l - 1);
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int _ = 1;
    // cin >> _;
    while (_--) {
        solve();
    }
    return 0;
}