#include <bits/stdc++.h>
using namespace std;
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	string s;
	long long k, i = 0;
	cin >> s >> k;
	i -= k;
	for (char c : s)
		if (!++i || c - '1')
			return cout << c << endl, 0;
}
### Prompt

Develop a solution in cpp to the problem described below:

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.

Examples

Input

1214
4


Output

2


Input

3
157


Output

3


Input

299792458
9460730472580800


Output

2

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	string s;
	long long k, i = 0;
	cin >> s >> k;
	i -= k;
	for (char c : s)
		if (!++i || c - '1')
			return cout << c << endl, 0;
}
```