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 ll = long long;
using namespace std;
#define rep(i,n) for(int i=0, i##_len=(int)(n); i<i##_len; i++)
#define reps(i,n) for(int i=1 , i##_len=(int)(n);i<=i##_len;i++)
#define rrep(i,n) for(int i=((int)(n)-1);i>=0;i--)
#define rreps(i,n) for(int i=((int)(n));i>0;i--)
#define repi(i,x) for(auto i=(x).begin(),i##_fin=(x).end();i!=i##_fin;i++)
#define all(x) (x).begin(), (x).end()
#define F first
#define S second
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define eb emplace_back
string solve(bool a) { return ((a) ? "Yes" : "No"); }
typedef vector<int> Vi;
typedef vector<Vi> VVi;
typedef pair<int , int> Pi;
typedef vector<Pi> VPi;
typedef vector<long long> V;
typedef vector<V> VV;
typedef pair<long long , long long> P;
typedef vector<P> VP;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1;} return 0;}
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1;} return 0;}
template <class T, class U>ostream& operator<<(ostream& os, const pair<T, U>& p) {
	os << "(" << p.first << "," << p.second << ")";
	return os;
}
template <class T>ostream& operator<<(ostream& os, const vector<T>& v) {
	os << "{";
	rep(i, v.size()) {
		if (i) os << ",";
		os << v[i];
	}
	os << "}";
	return os;
}
template <class T, class U>istream& operator>>(istream& is, pair<T, U>& p) {
	is >> p.first >>p.second ;
	return is;
}
template <class T>istream& operator>>(istream& is,  vector<T>& v) {
	rep(i, v.size()) {
		is >> v[i];
	}
	return is;
}
const long long INFLL = 1LL<<60;
const int INF = 1<<30;
const double PI=acos(-1);
int main(){
	string s;
	ll k;
	char ans=1;
	cin >>s>>k;
	for (int i = 0; i < k;i++){
		if(s[i]!='1'){
			cout << s[i] << endl;
			return 0;
		}
	} cout << 1<< endl;
}