Let's write all the positive integer numbers one after another from 1 without any delimiters (i.e. as a single string). It will be the infinite sequence starting with 123456789101112131415161718192021222324252627282930313233343536...

Your task is to print the k-th digit of this sequence.

Input

The first and only line contains integer k (1 ≤ k ≤ 10000) — the position to process (1-based index).

Output

Print the k-th digit of the resulting infinite sequence.

Examples

Input


7


Output


7


Input


21


Output


5
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, i, j, x, dig = 1;
  cin >> n;
  for (i = 1; i <= n; i++) {
    if (dig <= 9) {
      if (i == n) {
        x = dig;
        break;
      }
      dig++;
    } else if (dig > 9 && dig < 100) {
      if (i == n) {
        x = dig / 10;
        break;
      }
      i++;
      if (i == n) {
        x = dig % 10;
        break;
      }
      dig++;
    } else if (dig > 99 && dig < 1000) {
      if (i == n) {
        x = dig / 100;
        break;
      }
      i++;
      if (i == n) {
        x = (dig / 10) % 10;
        break;
      }
      i++;
      if (i == n) {
        x = dig % 10;
        break;
      }
      dig++;
    } else if (dig > 999 && dig < 10000) {
      if (i == n) {
        x = dig / 1000;
        break;
      }
      i++;
      if (i == n) {
        x = (dig / 100) % 10;
        break;
      }
      i++;
      if (i == n) {
        x = (dig / 10) % 10;
        break;
      }
      i++;
      if (i == n) {
        x = dig % 10;
        break;
      }
      dig++;
    } else if (dig == 10000) {
      if (i == n) {
        x = dig / 10000;
        break;
      }
      i++;
      if (i == n) {
        x = (dig / 1000) % 10;
        break;
      }
      i++;
      if (i == n) {
        x = (dig / 100) % 10;
        break;
      }
      i++;
      if (i == n) {
        x = (dig / 10) % 10;
        break;
      }
      i++;
      if (i == n) {
        x = dig % 10;
        break;
      }
      dig++;
    }
  }
  cout << x << endl;
  return 0;
}
