#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define all(x) x.begin(),x.end()
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define rep_r(i,a,b) for(int i=a;i>=b;i--)
#define each(a,x) for (auto& x : a)
using pi = pair<int,int>;
using pl = pair<ll,ll>;
using vi = vector<int>;
using vl = vector<ll>;
#define sz(x) int(x.size())
#define so(x) sort(all(x))
#define so_r(x) sort(all(x),greater<int>())
#define lb lower_bound
#define ub upper_bound
const char nl = '\n';
int dx[4] = {1,-1,0,0};
int dy[4] = {0,0,1,-1};
int bit_cnt(int x){
    return __builtin_popcount(x);
}
ll bex(ll a, ll b, ll mod = 1e9 + 7){ll res = 1LL; while(b){ if (b&1) res = res * a % mod; a = a * a % mod; b >>= 1;} return res;}
template<class t,class u> bool chmax(t&a,u b){if(a<b){a=b;return true;}else return false;}
template<class t,class u> bool chmin(t&a,u b){if(b<a){a=b;return true;}else return false;}

const int MOD = 998244353;
const int N = 5e5 + 5;
const ll INF = 1e16;
int n;
ll k; 
string s;
// https://yukicoder.me/problems/no/78
// brief: n icecreams in 1 box, 3 types of icecream: "truot", "trung 1 que", "trung 2 que"
// lay lan luot que kem theo thu tu, type cua que kem duoc ghi trong S (0 || 1 || 2)
// can mua it nhat bao nhieu que kem de co the an duoc it nhat k que kem

// n <= 50, k <= 2e9
pair<int,int> one_box(){
    int buy = 0; 
    int remain = 0;
    rep(i,1,n){
        buy++;
        int j = i;
        int prize = s[i-1] - '0'; 
        while (prize && j < n) {
            prize--;
            j++;
            if (j > i) prize += s[j-1] - '0';
        }
        i = j;
        remain = prize;
    }
    return {buy, remain};
}
int find1(int k){
    int buy = 0;
    if (k == 0) return 0;
    rep(i,1,n){ 
        buy++;
        int j = i;
        int prize = s[i-1] - '0'; 
        while (prize && j < n) {
            prize--; 
            j++; 
            if (j > i) prize += s[j-1] - '0';
        }
        i = j;
        if (i >= k) return buy;
    }
    return n;
}
void solve(){
    cin >> n >> k >> s; 
    pair<int,int> p = one_box(); 
    int buy = p.first, remain = p.second;
    // cout << buy << ' ' << remain << nl;
    if (k <= n) {
        cout << find1(k) << nl;
        return;
    }
    if (k > n && remain >= buy){
        cout << buy << nl;
        return;
    }

    // extra buy times we need to spend for full fill other box
    int next_buy = buy - remain; 

    ll boxes = k / n; // total full-boxes we need to buy 
    ll still = k % n; // number of icecream still need to buy

    ll ans = 1 * buy + next_buy * (boxes - 1) + max(find1(still) - remain, 0);
    cout << ans << nl;
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int t = 1; 
    // cin >> t;

    while (t--){
        solve();
    }
}