You are given three integers a, b and x. Your task is to construct a binary string s of length n = a + b such that there are exactly a zeroes, exactly b ones and exactly x indices i (where 1 ≤ i < n) such that s_i ≠ s_{i + 1}. It is guaranteed that the answer always exists.

For example, for the string "01010" there are four indices i such that 1 ≤ i < n and s_i ≠ s_{i + 1} (i = 1, 2, 3, 4). For the string "111001" there are two such indices i (i = 3, 5).

Recall that binary string is a non-empty sequence of characters where each character is either 0 or 1.

Input

The first line of the input contains three integers a, b and x (1 ≤ a, b ≤ 100, 1 ≤ x < a + b).

Output

Print only one string s, where s is any binary string satisfying conditions described above. It is guaranteed that the answer always exists.

Examples

Input

2 2 1


Output

1100


Input

3 3 3


Output

101100


Input

5 3 6


Output

01010100

Note

All possible answers for the first example: 

  * 1100; 
  * 0011. 



All possible answers for the second example: 

  * 110100; 
  * 101100; 
  * 110010; 
  * 100110; 
  * 011001; 
  * 001101; 
  * 010011; 
  * 001011. 
import java.util.*;

public class B
{
	public static void main(String[] args)
	{
		Scanner input = new Scanner(System.in);

		int a = input.nextInt();
		int b = input.nextInt();
		int x = input.nextInt();

		if(x == 1)
		{
			while(a-- > 0)
				System.out.print(0);
			while(b-- > 0)
				System.out.print(1);
			System.out.println();
			return;
		}

		boolean nextZero = false;
		if(a < b)
		{
			System.out.print(1);
			b--;
			nextZero = true;
		}
		else
		{
			System.out.print(0);
			a--;
		}

		while(x > 2)
		{
			if(nextZero)
			{
				System.out.print(0);
				a--;
				nextZero = false;
			}
			else
			{
				System.out.print(1);
				b--;
				nextZero = true;
			}
			x--;
		}

		if(nextZero)
		{
			while(a-- > 0)
				System.out.print(0);
			while(b-- > 0)
				System.out.print(1);
		}
		else
		{
			while(b-- > 0)
				System.out.print(1);
			while(a-- > 0)
				System.out.print(0);
		}
		System.out.println();

	}
}