Problem:
﻿Score: 300 points

Problem Statement

Mr. Infinity has a string S consisting of digits from 1 to 9. Each time the date changes, this string changes as follows:

Each occurrence of 2 in S is replaced with 22. Similarly, each 3 becomes 333, 4 becomes 4444, 5 becomes 55555, 6 becomes 666666, 7 becomes 7777777, 8 becomes 88888888 and 9 becomes 999999999. 1 remains as 1.

For example, if S is 1324, it becomes 1333224444 the next day, and it becomes 133333333322224444444444444444 the day after next.
You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days?

Constraints

S is a string of length between 1 and 100 (inclusive).

K is an integer between 1 and 10^{18} (inclusive).

The length of the string after 5 \times 10^{15} days is at least K.

Input

Input is given from Standard Input in the following format:

S
K

Output

Print the K-th character from the left in Mr. Infinity's string after 5 \times 10^{15} days.

Sample Input 1

1214
4

Sample Output 1

2

The string S changes as follows:

Now: 1214

After one day: 12214444

After two days: 1222214444444444444444

After three days: 12222222214444444444444444444444444444444444444444444444444444444444444444

The first five characters in the string after 5 \times 10^{15} days is 12222. As K=4, we should print the fourth character, 2.

Sample Input 2

3
157

Sample Output 2

3

The initial string is 3. The string after 5 \times 10^{15} days consists only of 3.

Sample Input 3

299792458
9460730472580800

Sample Output 3

2

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#pragma GCC optimize("Ofast")

using namespace std;
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
typedef long long ll;
#define rep(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define REP(i,num,n) for(int i=num, i##_len=(n); i<i##_len; ++i)
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
#define print(x) (cout << (x) << endl)
#define pb push_back
#define mp make_pair
#define sz(x) int(x.size())
template<class T> inline void Yes(T condition){ if(condition) cout << "Yes" << endl; else cout << "No" << endl; }
#define fs first
#define sc second
// 多倍長
#include <boost/rational.hpp>
#include <boost/multiprecision/cpp_int.hpp>
using boost::rational;
using boost::multiprecision::cpp_int;

const int INF = 1e9;
const ll LLINF = 1e16;
const ll MOD = 1e9+7;
template<class T> inline void add(T &a, T b){a = ((a+b) % MOD + MOD) % MOD;};

template<typename T>
vector<T> make_v(size_t a){return vector<T>(a);}
 
template<typename T,typename... Ts>
auto make_v(size_t a,Ts... ts){
  return vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...));
}
 

// 指定した文字cが文字列に何文字入ってるか
ll stringcount(string s, char c) {
    return count(s.cbegin(), s.cend(), c);
}


// 割り算の切り上げ処理
template<typename T>
T rp(T a, T b) {
  return (a + b - 1) / b;
}

// 桁和
template<typename T>
T digsum(T n) {
    T res = 0;
    while(n > 0) {
        res += n%10;
        n /= 10;
    }
    return res;
}

// 回文判定
bool kai (string s) {
    string t{s};
    reverse(begin(t), end(t));
    return s == t;
}

using pint = pair<int, int>;
using pll = pair<ll, ll>;


vector<string> split(const string &s, char delim) {
    vector<string> elems;
    string item;
    for (char ch: s) {
        if (ch == delim) {
            if (!item.empty())
                elems.push_back(item);
            item.clear();
        }
        else {
            item += ch;
        }
    }
    if (!item.empty())
        elems.push_back(item);
    return elems;
}


int ng[500];
int dp[500];

int main(void)
{
    cout.tie(0);
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);
    string s;
    cin >> s;
    ll k;
    cin >> k;
    int i = 0;
    while(s[i] == '1'){
        i++;
    }
    if(k <= i) print(1);
    else print(s[i]);

    

}