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>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<vvll> vvvll;
typedef vector<ld> vld;
typedef vector<string> vstr;
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;
typedef priority_queue<ll, vector<ll>, greater<ll>> spqll;  // 小さい順に取り出し
typedef priority_queue<ll, vector<ll>, less<ll>> bpqll;     // 大きい順に取り出し

#define REP(i, n) for (ll i = 0; i < (ll)(n); i++)
#define IREP(i, v) for (auto i = (v).begin(); i != (v).end(); ++i)
#define ALL(v) (v).begin(), (v).end()
#define endl "\n"

ll INF = 1e9;
ll MOD = 1000000007;
ll LINF = 1e18;
ld EPS = 1e-9;
ld PI = M_PI;
vll dx = {1, 0, -1, 0, 1, -1, -1, 1};
vll dy = {0, 1, 0, -1, 1, 1, -1, -1};
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a;} //最大公約数
ll lcm(ll a, ll b) { return a / gcd(a, b) * b;}     //最小公倍数

void yes(){ cout << "YES" << endl;}
void no(){ cout << "NO" << endl;}

//-----------------------------------------

//-----------------------------------------
int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(10);

  string s;
  cin >> s;
  ll k;
  cin >> k;

  ll i = 0;
  while(1){
    if(s[i] == '1'){
      k--;
      if(k == 0){
        cout << 1 << endl;
        return 0;
      }
    }
    else{
      cout << s[i] << endl;
      return 0;
    }
    i++;
  }





  return 0;
}