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 codeforces.Round494;

import java.util.Scanner;

public class BinaryStringConstructing {

    private static String construct(int zeros, int ones, int d) {
        StringBuilder sb = new StringBuilder();
        String left = (zeros > ones) ? "0":"1";
        int leftCount = (zeros > ones) ? zeros: ones;
        String right = (zeros <= ones) ? "0":"1";
        int rightCount = (zeros <= ones) ? zeros:ones;

        if (rightCount * 2 == d) {
            sb.append(left);
            leftCount--;
            while (rightCount > 0) {
                sb.append(right);
                sb.append(left);
                leftCount--;
                rightCount--;
            }
            while (leftCount > 0) {
                sb.append(left);
                leftCount--;
            }
            return sb.toString();
        } else {
            d--;
        }

        if (d % 2 == 1) {
            sb.append(right);
            sb.append(left);
            d--;
            leftCount--;
            rightCount--;
        } else {
            sb.append(left);
            leftCount -= 1;
        }
        while (d > 0) {
            sb.append(right);
            sb.append(left);
            leftCount--;
            rightCount--;
            d-=2;
        }
        while (leftCount > 0) {
            sb.append(left);
            leftCount--;
        }
        while (rightCount > 0) {
            sb.append(right);
            rightCount--;
        }

        return sb.toString();
    }
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int zeros = s.nextInt();
        int ones = s.nextInt();
        int d = s.nextInt();

        System.out.println(construct(zeros, ones, d));

    }
}
