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: Accepted)

#include <iostream>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <algorithm>
#include <utility>
#include <functional>
#include <fstream>
#define debug(x) cout << #x << " = " << x << endl
#define FOR(i,a,b) for(int i = (a); i < (b); i++)
#define rep(i,n) for(int i = 0; i < (n); i++)
#define repr(i,n) for(int i = (n); i > -1; i--)
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
const int MOD = 1000000007;
const double EPS = 1e-9;


int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);

  string s;
  ll k;
  cin>>s>>k;
  ll x;
  rep(i,s.size()) {
    if(s[i] != '1') {
      x = i;
      break;
    }
  }
  char ans;
  if(k < x + 1) {
    ans = '1';
  } else {
    ans = s[x];
  }
  cout << ans << endl;

  return 0;
}