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 B {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int a = sc.nextInt();
		int b = sc.nextInt();
		int x = sc.nextInt();
		int sum = a +b;
		int a1 = a;
		int b1 = b;
		int x1 = x;
		
		StringBuilder res = new StringBuilder();
		res.append("0");
		a--;
		int k = 1;
		while(x > 0){
			if(k % 2 == 1){
				res.append("1");
				b--;
			}else{
				res.append("0");
				a--;
			}
			k++;
			x--;
		}
		while(b > 0){
			res.insert(1, "1");
			b--;
		}
		while(a > 0){
			res.insert(0, "0");
			a--;
		}
		if(res.length() == sum){
			System.out.println(res.toString());
			return;
		}else{
			res = new StringBuilder();
			res.append("1");
			b1--;
			k = 1;
			while(x1 > 0){
				if(k % 2 == 1){
					res.append("0");
					a1--;
				}else{
					res.append("1");
					b1--;
				}
				k++;
				x1--;
			}
			while(a1 > 0){
				res.insert(1, "0");
				a1--;
			}
			while(b1 > 0){
				res.insert(0, "1");
				b1--;
			}
			System.out.println(res.toString());
		}
	}

}
