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.*;
import java.util.Locale;
import java.util.StringTokenizer;

public class B {
    String fileName = "<name>";

    public void solve() throws IOException {
        int b = nextInt();
        int a = nextInt();
        int x = nextInt();
        StringBuilder sb = new StringBuilder(a + b);
        if (b < a) {
            sb.append(1);
            a--;
        } else {
            sb.append(0);
            b--;
        }
        for (int i = 0; i < x; i++) {
            int val = 1 - Integer.parseInt(sb.substring(i, i + 1));
            sb.append(val);
            if (val == 1) {
                a--;
            } else {
                b--;
            }
        }
        int start = sb.indexOf("1");
        for (int i = 0; i < a; i++) {
            sb.insert(start, 1);
        }
        start = sb.indexOf("0");
        for (int i = 0; i < b; i++) {
            sb.insert(start, 0);
        }
        out.print(sb.toString());
    }

    public void run() {
        try {
            br = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);

            solve();

            out.close();
        } catch (IOException e) {
            e.printStackTrace();
            System.exit(1);
        }
    }

    BufferedReader br;
    StringTokenizer in;
    PrintWriter out;

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

    public int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    public double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    public long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    public static void main(String[] args) throws IOException {
        Locale.setDefault(Locale.US);
        new B().run();
    }
}