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. 

//package codeforce;

import java.util.Scanner;

public class b {
	public static void main(String[] args) {
    	Scanner sc = new Scanner(System.in);
        int ar[] = new int[2];
        ar[0] = sc.nextInt();
        ar[1] = sc.nextInt();
        int z = sc.nextInt();
        StringBuilder sb = new StringBuilder();
        int s = 0;
        if(ar[0]<ar[1])
        s=1;
		String res = "";
        int x = s;
        int y = s^1;
        if(z%2==0){
            while(z-->0){
				res = res + s;
                ar[s]--;
                s = s^1;
            }
            while(ar[x]-->1)
				res = x + res;
            while(ar[y]-->0)
				res = res + y;
 
			res = res + x;
        }
        else{
            z++;
            while(z-->0){
				res = res + s;
                ar[s]--;
                s = s^1;
            }
            while(ar[x]-->0){
				res = x + res;
            }
            while(ar[y]-->0){
				res = res + y;
            }
        }
 
		System.out.println(res);
        
	}
}
