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 fi=new Scanner(System.in);
		int a=fi.nextInt();
		int b=fi.nextInt();
		int x=fi.nextInt();
		String s="";
		int flag=0;
		if(x%2==0)
		{
			//System.out.println("la3");

			if(a>=b)
			{
				flag=1;
			}
			else
			{
				s+="1";
				--b;
			}
			--x;
		}


		if(x%2==1)
		{
			String str="";
			String str1="";
			String str2="";
			int loo=(x+1)/2;
			for(int i=0;i<loo;i++)
			{
				str+="01";
					--a;
				    --b;
				//System.out.println("la2");
			}
			for(int i=0;i<a;i++)
			{
				str1+="0";
				//.out.println("la");
			}
			for(int i=0;i<b;i++) {
				str2 += "1";
				//System.out.println("la1");
			}
			s+=str1+str+str2;
			if(flag==1)
			{
				s=s.substring(1);
				s+="0";
			}
			System.out.println(s);
			return ;

		}
		fi.close();

	}
}
