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;

#define ll long long
#define ld long double
#define P pair<ll, ll>

#define FORm(i, m) for (auto i = m.begin(); i != m.end(); i++)

#define all(v) v.begin(), v.end()

#define sortAl(v) sort(all(v))
#define sortAlr(v)                                                             \
  sort(v.begin(), v.end());                                                    \
  reverse(v.begin(), v.end())

#define cout(n) cout << fixed << setprecision(n)

#define rep(i, n) for (int i = 0; i < n; i++)
#define REP(i, n) for (int i = 1; i < n; i++)

template <class T> inline void chmax(T &a, T b) { a = std::max(a, b); }
template <class T> inline void chmin(T &a, T b) { a = std::min(a, b); }

#define mod (int)(1e9 + 7)
#define INF LLONG_MAX

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

  string s;
  cin >> s;

  ll k;
  cin >> k;

  int index = 0;
  FORm(it, s) {
    if (*it != '1') {
      cout << *it << endl;
      return 0;
    }
    index++;
    if (index >= k) {
      break;
    }
  }

  cout << 1 << endl;
  return 0;
}