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.*;
public class MyClass {
    public static void main(String args[]) {
        Scanner sc= new Scanner(System.in);
        int a,b,c;
        a=sc.nextInt();
        b=sc.nextInt();
        c=sc.nextInt();
        boolean f ; String str ="";
        if(a>b) f=true; else f= false;
        while(c-->-1){
            if(f) str+=0; else str+=1;
            if(f) a--; else b--;
            f=!f;
        }
        if(a>0){
            String temp ="";
            for(int i=0;i<a;i++) temp+=0;
            int ai = str.indexOf('0');
            str = str.substring(0,ai)+temp+str.substring(ai,str.length());
        }
        if(b>0){
            String temp ="";
            for(int i=0;i<b;i++) temp+=1;
            int bi = str.indexOf('1');
            str = str.substring(0,bi)+temp+str.substring(bi,str.length());
        }
        System.out.println(str);
    }
}
