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
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Washoum
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        inputClass in = new inputClass(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        ADigitsSequenceEasyEdition solver = new ADigitsSequenceEasyEdition();
        solver.solve(1, in, out);
        out.close();
    }

    static class ADigitsSequenceEasyEdition {
        public void solve(int testNumber, inputClass sc, PrintWriter out) {
            StringBuilder s = new StringBuilder();
            int i = 1;
            while (s.length() < 10001) {
                s.append(i);
                i++;
            }
            int k = sc.nextInt();
            out.println(s.charAt(k - 1));
        }

    }

    static class inputClass {
        BufferedReader br;
        StringTokenizer st;

        public inputClass(InputStream in) {

            br = new BufferedReader(new InputStreamReader(in));
        }

        public String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }
}

