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 static java.lang.Math.*;
import java.io.*;

public class SolutionB {
       
      public static void main(String args[])throws IOException{
              Scanner sc = new Scanner(System.in);
              int a = sc.nextInt();
              int b = sc.nextInt();
              int k = sc.nextInt();
              Deque<Integer> que = new LinkedList<Integer>();
              if(a >= b){
                     que.offer(0);
                     a--;
              }else{
                     que.offer(1);
                     b--;
              }
              
              while(a > 0 && b > 0 && k > 1){
                     if(que.peekLast() == 1){
                            que.offer(0);
                            a--;
                     }else{
                            que.offer(1);
                            b--;
                     }
                     k--;
              }
              if(que.peekLast() == 1){
                     while(b > 0){
                            que.offer(1);
                            b--;
                     }
                     while(a > 0){
                            que.offer(0);
                            a--;
                     }
              }else{
                     while(a > 0){
                            que.offer(0);
                            a--;
                     }
                     while(b > 0){
                            que.offer(1);
                            b--;
                     }
              }
              while(!que.isEmpty())
                     System.out.print(que.poll());
              System.out.println();
      }
}