#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;

long long n,m,k;
long long pre[N];//前缀中0的数量

long long dp[N][6][2];
string s;
bool check(int mid)
{
    for (int i = 0; i <= n;i++)
    {
        for(int j = 0;j<=k;j++)
        {
            dp[i][j][0]=dp[i][j][1]=0x3f3f3f3f;
        }
    }
    dp[0][0][0] = 0;

    for(int i = 1;i<=n;i++)
    {
        for(int j = 0;j<=k;j++)
        {

            if(s[i]=='0')
            {
				dp[i][j][0] = min({dp[i][j][0], dp[i-1][j][0], dp[i-1][j][1]});
				dp[i][j][1] = min({dp[i][j][1], dp[i-1][j][1] + 1});
            }
            else
            {
				dp[i][j][0] = min(dp[i][j][0], dp[i-1][j][0]);
				dp[i][j][1] = min(dp[i][j][1], dp[i-1][j][1]);
            }


			if (i >= mid && j >= 1)
            {
                dp[i][j][1] = min((dp[i][j][1]), (dp[i-mid][j-1][0] + pre[i] - pre[i-mid]));
            }
				

        }
    }

return min(dp[n][k][0], dp[n][k][1]) <= m;



}





int main()
{
    cin >> n >> m >> k;


    cin >> s;
    s = ' ' + s;

    for(int i =1;i<= n;i++)
    {
        pre[i] += (pre[i - 1] + s[i] == '0');
    }


    int l=0,r=2e5;
    while (l<r)
    {
        int mid = (l + r + 1) >> 1;
        if(check(mid))l=mid;
        else
            r = mid - 1;

    }
    
    if(l)
        cout << l << '\n';
    else
        cout << "-1\n";

    return 0;
}