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.util.Scanner;


public class S448D {

	public long solve(long n, long m, long k) {
		if (k == 1)
			return 1;
		else if (k == n * m)
			return n * m;
		else {
			long left = 0;
			long right = n * m;
			long count = 0;
			
			while (left != right) {
				long mid = (left + right) / 2;
				count = 0;
				for (long i = 1; i <= n; i ++) {
					count += Math.min(mid / i, m);
				}
				if (count < k) {
					left = mid + 1;
				} else {
					right = mid;
				}
			}
			return left;
		}
	}
	
	public static void main(String[] args) {
		S448D s448d = new S448D();
		/*
		int n0 = 2;
		int m0 = 2;
		int k0 = 2;
		int n1 = 2;
		int m1 = 3;
		int k1 = 4;
		int n2 = 1;
		int m2 = 10;
		int k2 = 5;
		int n3 = 500000;
		int m3 = 500000;
		int k3 = 500000;
		System.out.println(s448d.solve(n0, m0, k0));
		System.out.println(s448d.solve(n1, m1, k1));
		System.out.println(s448d.solve(n2, m2, k2));
		System.out.println(s448d.solve(n3, m3, k3));
		*/
		
		Scanner in = new Scanner(System.in);
		long n = in.nextInt();
		long m = in.nextInt();
		long k = in.nextLong();
		
		System.out.println(s448d.solve(n, m, k));
	}

}
