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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define REP(i,n) for (int i=0;i<(n);i++)
using namespace std;
typedef long long ll;
int main() {
    string S;
    cin >> S;
    ll K; cin >> K;
    int N = S.length();
    vector<pair<int, ll>> s(N);
    REP(i, N){
        s[i].first = S[i] - '0';
        s[i].second = 1;
    }
    ll sum = 0;
    while(sum < 8){
        REP(i,N){
            s[i].second = s[i].first*s[i].second;
        }
        sum++;
    }
    // REP(i, S.length()) cout << s[i].first << " " << s[i].second << endl;
    ll count = 0;
    REP(i,N){
        REP(j,s[i].second){
            count += 1;
            if(count == K){
                cout << s[i].first;
                return 0;
            }
        }
    }
    return 0;
}