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 <bits/stdc++.h>

using namespace std;
typedef long long ll;

#define MOD (long long)(1e9+7)
#define REP(i,n) for(ll i = 0; i < n; i++)
#define REP1(i,n) for(ll i = 1; i <= n; i++)

ll gcd(ll a, ll b)
{
  if(b == 0) return a;
  return gcd(b, a % b);
}

int main(void)
{
  string s; cin >> s;
  ll k; cin >> k;
  // ll x, y, z; cin >> x >> y >> z;
  // vector<ll> a(n); REP(i, n) cin >> a[i];
  char ans;
  // bool ok = false;
  ll l;
  for(l = 0; s[l] == '1'; l++) {};
  if(k <= l) {
    ans = '1';
  } else {
    ans = s[l];
  }

  cout << ans << endl;
  // cout << (ok?"Yes":"No") << endl;
}