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<vector>
#include<algorithm>
#include<cctype>
#include<utility>
#include<string>
#include<cmath>
#include<cstring>
#include<queue>
#include<map>
#include<set>
using namespace std;

#define REP(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORR(i, m, n) for(int i = m;i >= n;i--)
#define VSORT(v) sort(v.begin(), v.end());
#define pb(a) push_back(a)


typedef long long ll;

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const int INF = 1e9 + 7;

#define ARRAY_MAX 100005

template<typename T>
vector<T> make_v(size_t a) { return vector<T>(a); }
template<typename T, typename... Ts>
auto make_v(size_t a, Ts... ts) {
	return vector<decltype(make_v<T>(ts...))>(a, make_v<T>(ts...));
}
template<typename T, typename V>
typename enable_if<is_class<T>::value == 0>::type
fill_v(T& t, const V& v) { t = v; }
template<typename T, typename V>
typename enable_if<is_class<T>::value != 0>::type
fill_v(T& t, const V& v) {
	for (auto& e : t) fill_v(e, v);
}


int main() {

	string s;
	cin >> s;
	ll k;
	cin >> k;
	
	ll index = 0;
	char c = s[0];

	bool flag = false;
	for (int i = 0; i < s.size(); i++)
	{
		if (s[i] == '1')
		{
			index++;
		}
		else
		{
			flag = true;
			c = s[i];
			break;
		}
	}
	if (!flag)
	{
		c = s[s.size() - 1];
	}

	if (index >= k)
	{
		cout << s[k-1] << endl;
	}
	else
	{
		cout << c << endl;
	}
	
	return 0;
}