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 <iostream>
#include <string> 
#include <vector> 
#include <algorithm> 
#include <utility> 
#include <tuple> 
#include <cstdint> 
#include <cstdio> 
#include <map> 
#include <queue> 
#include <set> 
#include <stack> 
#include <deque> 
#include <unordered_map> 
#include <unordered_set> 
#include <bitset> 
#include <cctype> 
#include <math.h>
#include <cmath>
using namespace std;

#define int long long
#define endl "\n"
#define all(v)  v.begin(),v.end()
#define fir first
#define sec second
#define fro front
#define m_p make_pair
#define rep(i,n)  for(int i=0; i<(int) (n); i++)
#define vecin(v) for(int i=0; i<(int)v.size(); i++)cin>>v[i];

int mod;
const int INF = 2000000015;
const int MAX = 510000;
const int MOD = 1000000007;
long long fac[MAX], finv[MAX], inv[MAX];
int jousu(int x00, int y00) {
	int z00 = 1;
	for (int i = 0; i < y00; i++) {
		z00 *= x00;
	}
	return z00;
}
int keta(int x00) {
	int z00 = x00;
	int w00 = 0;
	while (z00 != 0) {
		z00 /= 10;
		w00++;
	}
	return w00;
}
int krag(double xx) {
	return INF - (int)(INF - xx);
}


int modinv(int a, int m) {
	int b = m, u = 1, v = 0;
	while (b) {
		int t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}
int modpow(int a, int n) {
	int res = 1;
	while (n > 0) {
		if (n & 1) res=res* a% mod;
		a = a * a % mod;
		n >>= 1;
	}
	return res;
}
int com(int xx, int yy) {
	int zz = 1;
	for (int i = xx; i > xx - yy; i--) {
		zz *= i;
		zz %= mod;
	}
	for (int i = 1; i <= yy; i++) {
		zz *= modinv(i, mod);
		zz %= mod;
	}
	return zz;
}
int gcd(int xx, int yy) {
	int p = xx;
	int q = yy;
	if (q > p)swap(p, q);
	while (p % q != 0) {
		p %= q;
		swap(p, q);
	}
	return q;
}
int lcm(int xx, int yy) {
	return xx * yy / gcd(xx, yy);
}
bool prime(int xx) {
	int a = xx;
	for (int i = 2; i * i <= xx; i++) {
		if (xx % i == 0) {
			return 0;
		}
	}
	return 1;
}


signed main() {
	string s;
	int k;
	cin >> s >> k;
	rep(i, s.size()) {
		if (s[i]-'0' > 1 || k - 1 == i) {
			cout << s[i]-'0' << endl;
			return 0;
		}
	}
}