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;
bool isvowel(char c) {
  if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' || c == 'A' ||
      c == 'E' || c == 'I' || c == 'O' || c == 'U')
    return true;
  return false;
}
bool prime[1000001];
int spf[1000001];
vector<int> primes;
void sieve() {
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= 1000000; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= 1000000; i += p) {
        prime[i] = false;
        spf[i] = p;
      }
    }
  }
  for (int p = 2; p <= 1000000; p++)
    if (prime[p]) primes.push_back(p);
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int fast_exp(long long int x, long long int n) {
  long long int ans = 1;
  while (n) {
    if (n & 1) ans *= x;
    n = n >> 1;
    x = x * x;
  }
  return ans;
}
long long int mod_exp(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int main() {
  string s;
  int c = 1;
  while (s.size() <= 10000) {
    ostringstream str1;
    str1 << c;
    string geek = str1.str();
    s += geek;
    c++;
  }
  int k;
  cin >> k;
  cout << s[k - 1];
}
