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.*;
import java.io.*;

public class R494D3N2 {
	public static void main(String[] args) throws Exception{
 		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
		String []a = bf.readLine().split(" ");
		int a1 = Integer.parseInt(a[0]);
		int b1 = Integer.parseInt(a[1]);
		int x = Integer.parseInt(a[2]);
		String[] ans = new String[x+1];
		int count0 = 0;
		int count1 = 0;
		if (a1 > b1){
			for(int i = 0;i<x+1;i++){
				if (i%2 == 0){
					ans[i] = "0";
					count0+=1;
				}
				else{
					ans[i] = "1";
					count1+=1;
				}
			}
		}
		
		else{
			for(int i = 0;i<x+1;i++){
				if (i%2 == 0){
					ans[i] = "1";
					count1+=1;
				}
				else{
					ans[i] = "0";
					count0+=1;
				}
			}
		}
		
		int k1 = a1-count0;
		int k2 = b1-count1;
		String addon1 = "";
		String addon0 = "";
		for(int i = 0;i<k1;i++){
			addon0 += "0";
		}
		for(int i = 0;i<k2;i++){
			addon1 += "1";
		}
		
		if (ans[0] == "0"){
		ans[0] += addon0;
		ans[1] += addon1;
		}
		else{
			ans[0] += addon1;
			ans[1] += addon0;
		}
		
		for(int i = 0;i<ans.length;i++){
			System.out.print(ans[i]);
		}
		
		
	}
}