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 java.io.*;
public class B{
	static long n;
	public static void main(String[] args) {
		MyScanner scan = new MyScanner();
		PrintWriter pw = new PrintWriter(System.out);
		int a = scan.nextInt();
		int b = scan.nextInt();
		int x = scan.nextInt();
		int arr[] = new int[a+b];
		StringBuilder str = new StringBuilder();
		int start = 0;
		int initial = 0;
		if(a < b)
		{
			start = 1;
		}
		initial = start;
		for(int i = 1;i<=x;i++)
		{
			if(start == 0)
				a--;
			else
				b--;
			//System.out.println(start+"fda i"+i);
			str.append(start+"");
			start = 1-start;
		}
		//pw.println(str+"fdasfd");
		if(initial == 0 && x %2 == 0)
		{
			pw.print(str);
			while(b-->0)
				pw.print(1);
			while(a-->0)
				pw.print(0);
		}
		else if(initial == 0 && x%2 == 1)
		{
			pw.print(str);
			while(a-->0)
				pw.print(0);
			while(b-->0)
				pw.print(1);
		}
		else if(initial == 1 && x%2 == 0)
		{
			pw.print(str);
			while(a-->0)
				pw.print(0);
			while(b-->0)
				pw.print(1);
		}
		else
		{
			pw.print(str);
			while(b-->0)
				pw.print(1);
			while(a-->0)
				pw.print(0);
		}
		pw.println();
		pw.close();
	}
}


//-------------------------------------------------------------------------------------------------------
class MyScanner {
    BufferedReader br;
    StringTokenizer st;

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

    public int mod(long x) {
        // TODO Auto-generated method stub
        return (int) x % 1000000007;
    }

    public int mod(int x) {
        return x % 1000000007;
    }

    boolean hasNext() {
        if (st.hasMoreElements())
            return true;
        try {
            st = new StringTokenizer(br.readLine());
        } catch (IOException e) {
            e.printStackTrace();
        }
        return st.hasMoreTokens();
    }

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

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

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

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

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