Bizon the Champion isn't just charming, he also is very smart.

While some of us were learning the multiplication table, Bizon the Champion had fun in his own manner. Bizon the Champion painted an n × m multiplication table, where the element on the intersection of the i-th row and j-th column equals i·j (the rows and columns of the table are numbered starting from 1). Then he was asked: what number in the table is the k-th largest number? Bizon the Champion always answered correctly and immediately. Can you repeat his success?

Consider the given multiplication table. If you write out all n·m numbers from the table in the non-decreasing order, then the k-th number you write out is called the k-th largest number.

Input

The single line contains integers n, m and k (1 ≤ n, m ≤ 5·105; 1 ≤ k ≤ n·m).

Output

Print the k-th largest number in a n × m multiplication table.

Examples

Input

2 2 2


Output

2


Input

2 3 4


Output

3


Input

1 10 5


Output

5

Note

A 2 × 3 multiplication table looks like this:
    
    
      
    1 2 3  
    2 4 6  
      
    


import java.io.*;
import java.sql.SQLSyntaxErrorException;
import java.util.*;
public class Question {
    static class Parser{
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;
        public Parser(InputStream in){
            din = new DataInputStream(in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0; }
        public long nextLong() throws Exception {
            long ret = 0;
            byte c = read();
            while (c <= ' ') c = read();
            boolean neg = c == '-';
            if (neg) c = read();
            do { ret = ret * 10 + c - '0';
                c = read();
            } while (c > ' ');
            if (neg) return -ret;
            return ret; }
        //reads in the next string
        public String next() throws Exception{
            StringBuilder ret = new StringBuilder();
            byte c = read();
            while (c <= ' ') c = read();
            do {
                ret = ret.append((char)c);
                c = read();
            } while (c > ' ');
            return ret.toString(); }
        public int nextInt() throws Exception {
            int ret = 0;
            byte c = read();
            while (c <= ' ') c = read();
            boolean neg = c == '-';
            if (neg) c = read();
            do {
                ret = ret * 10 + c - '0';
                c = read(); } while (c > ' ');
            if (neg) return -ret;
            return ret; }
        private void fillBuffer() throws Exception{
            bytesRead = din.read(buffer,
                    bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1) buffer[0] = -1; }
        private byte read() throws Exception {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++]; }}

static long ans = -1;
    static void binary(long l , long h, long N ,long K, long M){
        if(l>h)
            return;
        long mid = (l+h)/2l;
        long count = 0l ;
        for(long i = 1l;i<=N;i=i+1l){
            long val = mid/i;
            long min =Math.min(val,M);
            count+=min;

        }
        if(K>count){
            binary(mid+1l,h,N,K,M);}
            else{
                ans = mid;
                binary(l,mid-1l,N,K,M);
        }
    }
    public static void main(String args[]) throws Exception {
        Parser s=new Parser(System.in);
        PrintWriter out =new PrintWriter(System.out);
        long N = s.nextLong();
        long M = s.nextLong();
        long K = s.nextLong();
        binary(1l,N*M,N,K,M);
        System.out.println(ans);

    }
}
