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 Solution {
	public static void main(String[] args) {
		Solution solution = new Solution();
		solution.solve();
	}

	private void solve() {
		Scanner in = new Scanner(System.in);
		int b = in.nextInt();
		int a = in.nextInt();
		int x = in.nextInt();

		StringBuilder sb = new StringBuilder();

		for (int i = 0; i < (x + 1) / 2; ++i) sb.append("01");
		a -= (x + 1) / 2;
		b -= (x + 1) / 2;
		x -= ((x + 1) / 2) * 2 - 1;

		StringBuilder prefix = new StringBuilder();
		for (int i = 0; i < b; ++i) prefix.append('0');
		for (int i = 0; i < a; ++i) sb.append('1');

		if (x > 0) {
			if (b > 0) {
				prefix.setLength(prefix.length() - 1);
				sb.append('0');
			}
			else {
				sb.setLength(sb.length() - 1);
				if (prefix.length() > 0) prefix.setCharAt(0, '1');
				else prefix.append('1');
			}
		}
		System.out.println(prefix.toString() + sb.toString());
	}
}
