#include<bits/stdc++.h>
using namespace std;
const int N = 200001;
char v[N];
int sum[N];
int dp[N][6][2];//以i位置结尾,前面维护j个长度至少为k的最小花费
int n, m, k;
void init(int n)
{
    for (int i = 0; i <= n; i++)for (int j = 1; j <= 5; j++)
    {
        dp[i][j][0]=dp[i][j][1] = 1e9;
    }
}
bool check(int mid)
{
    init(n);
   // mid = 1;
    for (int i = 1; i <= n; i++)for (int j = 1; j <= k; j++)
    {
        if (v[i] == '1')
        {
            dp[i][j][1] = min(dp[i - 1][j][1],dp[i-1][j][0]);
            if ((i == n || v[i + 1] == '0') && i - mid  >= 0)
            {
                dp[i][j][1] = min(dp[i][j][1], dp[i - mid][j - 1][0] + (sum[i] - sum[i - mid]));
            }

        }
        else
        {
            dp[i][j][0] = min(dp[i - 1][j][0],dp[i-1][j][1]);
            if (v[i + 1] == '1')continue;//下个位置是1
            if (i - mid >= 0)
            {
               //cout << dp[i - mid][j - 1][0] + sum[i] - sum[i - mid] << endl;
                dp[i][j][1] = min(dp[i][j][1], dp[i - mid][j - 1][0] + sum[i] - sum[i - mid]);
            }
        }
       //cout << "以" << i << "位置1结尾前面有" << j << "段长度大于等于" << mid << "的最小花费" << dp[i][j][1] << endl;
        //cout << "以" << i << "位置0结尾前面有" << j << "段长度大于等于" << mid << "的最小花费" << dp[i][j][0] << endl;
        
    }
    return dp[n][k][0] <= m||dp[n][k][1]<=m;
}
signed main()
{
    cin >> n >> m >> k;
    for (int i = 1; i <= n; i++)
    {
        cin >> v[i];
    }
    for (int i = 1; i <= n; i++)
    {
        sum[i] = sum[i - 1];
        if (v[i] == '0')sum[i]++;
    }
    int ans = -1;
    int l = 1; int r = n;
    while (l <= r)
    {
        int mid = (l + r) / 2;
        if (check(mid))
        {
            ans = mid;
            l = mid + 1;
        }
        else
        {
            r = mid - 1;
        }
    }
    cout << ans << endl;
}