#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define modd(a,b) a=a%b
#define pb push_back
#define db double
#define lowbit(x) x&(-x)
#define cerr(x) cout<<#x<<"="<<x<<endl
const ll mod=1e9+7;
ll ksm(ll a,ll b) {ll ans=1,bs=a;while(b) {if(b&1) ans=ans*bs%mod;bs=bs*bs%mod;b>>=1;}return ans;}
#define fi first
#define se second
#define N 200010
int n,m,k;
int c[N],f[N][6],cnt[N];
bool is[N];
int fly[N];
bool ck(int mid) {
    for(int i=0;i<=n;i++) for(int j=0;j<=k;j++) f[i][j]=0x3f3f3f3f;
    for(int i=0;i<=n;i++) f[i][0]=0;
    for(int i=mid;i<=n;i++) {
        for(int j=1;j<=k;j++) {
            f[i][j]=f[i-1][j];
            if(c[i] && !is[i]) continue;
            int x=i-mid+1;
            if(c[x]) x=fly[x];
            else if(c[x-1]) x=fly[x-1];
            if(x==1) {//既然是要连成一个连续的1段那么就不是单纯的填1使得达到mid个了,而是要连续的话只能是枚举的起到到终点的所有数都填上1才行,所以答案的计算和mid就无关
                if(j==1) {
                    f[i][j]=min(f[i][j],i-cnt[i]);
                }
            }
            else {
                f[i][j]=min(f[i][j],f[x-2][j-1]+(i-x+1)-(cnt[i]-cnt[x-1]));
            }
        }
    }
    return f[n][k]<=m;
}
void solve() {
    cin>>n>>m>>k;
    for(int i=1;i<=n;i++) {char x;cin>>x;c[i]=x-'0';cnt[i]=cnt[i-1]+(c[i]==1);}
    for(int i=1;i<=n;i++) {
        if(c[i] && (i==n || c[i+1]==0)) is[i]=1;
    }
    for(int i=1;i<=n;i++) {
        if(!c[i]) continue;
        fly[i]=i;
        if(i>1 && c[i-1]) fly[i]=fly[i-1];
    }
    int l=1,r=n,mid,ans=-1;
    while(l<=r ) {
        mid=l+r>>1;
        if(ck(mid)) ans=mid,l=mid+1;
        else r=mid-1;
    }
    cout<<ans<<endl;
}
int main() {
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    // int T;
    // cin>>T;
    // while(T--) 
    solve();
    return 0;
}