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.Scanner;

public class BinaryStringConstructing {
	static Scanner sc;
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		sc=new Scanner(System.in);
		int a=sc.nextInt();
		int b=sc.nextInt();
		int c=sc.nextInt();
		String s="";
		char k='0';
		char j='1';
		if(b>a) {
			int temp=b;
			b=a;
			a=temp;
			k='1';
			j='0';
		}
		int i=0;
		if(c!=1) {
			for(i=0;i<c;i++) {
				if(i%2==0) {
					s+=k;
					a--;
				}
				else {
					s+=j;
					b--;
				}
			}
		}
		
		
		if(i%2==1) {
			for(i=0;i<a;i++) {
				s+=k;
			}
			for(i=0;i<b;i++) {
				s+=j;
			}
		}
		else {
			for(i=0;i<b;i++) {
				s+=j;
			}
			for(i=0;i<a;i++) {
				s+=k;
			}
		}
		System.out.println(s);
	}
}
