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 repd(i,a,b) for(int i=a;i<b;i++)
#define repds(i,a,b) for(int i=a+1;i<=b;i++)
#define rep(i,N) repd(i,0,N)
#define reps(i,N) repds(i,0,N)
#define debug(x) cout<<#x<<"="<<x<<endl
#define debugarr(arr,N,M){rep(i,N+1){rep(j,M+1){if(arr[i][j]==INF)printf(" INF");\
	else printf(" %3d",arr[i][j]);} cout<<endl;}}
#define pb push_back
#define pob pop_back
#define cint(a) int a;cin>>a
#define cint2(a,b) int a,b;cin>>a>>b
#define cint3(a,b,c) int a,b,c;cin>>a>>b>>c
typedef long long ll;
typedef pair<int,int> P;
const int INF=100000000;
const int dx[4]={ 0, 1, 0,-1};
const int dy[4]={ 1, 0,-1, 0};
//--------------------------------------//
string s;
int a[100];
ll k;
//10^18
//10^15*5.0
int main(){
	cin>>s;
	int z=s.size();
	rep(i,z) {
		a[i]=s[i]-'0';
	}
	cin>>k;
	ll oc=0;
	rep(i,z){
		if(a[i]!=1)
			break;
		oc++;
	}
	if(k<=oc) {
		cout<<"1"<<endl;
		return 0;
	}
	for(int i=0;;i++){
		if(a[i]!=1){
			cout<<a[i]<<endl;
			break;
		}
	}
	return 0;
}