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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.*;

public class binaryStringConstruction {
	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}

	public static void main(String[] args) {
		FastReader sc = new FastReader();
		
		int a=sc.nextInt();
		int b=sc.nextInt();
		int x=sc.nextInt();
		int one=0;
		int zero=0;
		int count=0;
		String s="";
		for(int i=0;i<a+b;i++)
		{
			if(b<a)
			{
				s=s+'0';
				s=s+'1';
			}
			else
			{
				s=s+'1';
				s=s+'0';
			}
		}
		//System.out.println(s);
		String s1="";
		if(x>1)
		{
			for(int i=0;i<s.length()-1;i++)
			{
				if(s.charAt(i)!=s.charAt(i+1))
				{
					
					s1=s1+s.charAt(i);
					count++;
					if(count==x-1)
					{
						s1=s1+s.charAt(i+1);
						break;
					}
				}
			}
		
	
			for(int i=0;i<s1.length();i++) 
			{
				if(s.charAt(i)=='0')
				{
					zero++;
				}
				else
				{
					one++;
				}
			}
			//System.out.println(s1);
			if(s1.charAt(s1.length()-1)=='0')
			{
				for(int i=zero;i<a;i++)
				{
					s1=s1+'0';
				}
				
				for(int i=one;i<b;i++)
				{
					s1=s1+'1';
				}
			}
			
			else if(s1.charAt(s1.length()-1)=='1')
			{
				for(int i=one;i<b;i++)
				{
					s1=s1+'1';
				}
				
				for(int i=zero;i<a;i++)
				{
					s1=s1+'0';
				}
			}
			
			System.out.println(s1);
		}
		
		else if(x==1)
		{
			for(int i=0;i<a;i++)
			{
				s1=s1+'0';
			}
			for(int i=0;i<b;i++)
			{
				s1=s1+'1';
			}
			System.out.println(s1);
		}
	}
}