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.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.StringTokenizer;

/**
 * Created by artur on 18/06/18
 */
public class A
{
    private BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    private StringTokenizer tokenizer = null;

    long mod = 1000000000;

    public static void main(String[] args) throws IOException
    {
        new A().solve();
    }

    private void solve() throws IOException
    {
        int a = ni();
        int b = ni();
        int x = ni();

        int c;
        StringBuilder sb = new StringBuilder();
        if (a > b) {
            c = 0;
        } else {
            c = 1;
        }

        for (int i = 0; i <= x; i++)
        {
            sb.append(c);
            if (c == 0)
            {
                a--;
            } else
            {
                b--;
            }
            c = 1 - c;
        }
        String res = sb.toString();
        StringBuilder bs = new StringBuilder();
        StringBuilder as = new StringBuilder();
        for (int i = 0; i < a; i++)
        {
            as.append('0');
        }
        for (int i = 0; i < b; i++)
        {
            bs.append('1');
        }
        res = res.substring(0, res.indexOf('1')) + bs.toString() + res.substring(res.indexOf('1'));
        res = res.substring(0, res.indexOf('0')) + as.toString() + res.substring(res.indexOf('0'));
        System.out.println(res);
    }


    void debug(Object... os)
    {
        System.out.println(Arrays.deepToString(os));
    }

    int ni() throws IOException
    {
        return Integer.parseInt(ns());
    }

    long nl() throws IOException
    {
        return Long.parseLong(ns());
    }

    double nd() throws IOException
    {
        return Double.parseDouble(ns());
    }

    String ns() throws IOException
    {
        while (tokenizer == null || !tokenizer.hasMoreTokens())
            tokenizer = new StringTokenizer(br.readLine());
        return tokenizer.nextToken();
    }

    String nline() throws IOException
    {
        tokenizer = null;
        return br.readLine();
    }

    private static long gcd(long a, long b)
    {
        while (b > 0)
        {
            long temp = b;
            b = a % b; // % is remainder
            a = temp;
        }
        return a;
    }

    private static long lcm(long a, long b)
    {
        return a * (b / gcd(a, b));
    }
}
