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.*;
public class d {
    public static void main(String args[]) throws IOException {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader sc = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        int a = sc.nextInt(), b = sc.nextInt(), x = sc.nextInt() + 1;
        int n = a + b;
        int[] count = {a, b};
        char[] c = {'0', '1'};
        int p = a > b ? 0 : 1;
        StringBuilder ans = new StringBuilder();
        for (int i = 0; i < n && a > 0 && b > 0; i++) {
            count[p]--;
            ans.append(c[p]);
            x--;
            if (x == 1)
                break;
            p = (p + 1) % 2;

        }
        for (int i = 0; i < count[p]; i++) {
            ans.append(c[p]);
        }
        p = (p + 1) % 2;

        for (int i = 0; i < count[p]; i++) {
            ans.append(c[p]);
        }
        out.print(ans);

        out.close();
    }
    public static void shuffle(long[] arr) {
        int n = arr.length;
        Random rnd = new Random();
        for (int i = 0; i < n; ++i) {
            long tmp = arr[i];
            int randomPos = i + rnd.nextInt(n - i);
            arr[i] = arr[randomPos];
            arr[randomPos] = tmp;
        }
    }
    public static void shuffle(int[] arr) {
        int n = arr.length;
        Random rnd = new Random();
        for (int i = 0; i < n; ++i) {
            int tmp = arr[i];
            int randomPos = i + rnd.nextInt(n - i);
            arr[i] = arr[randomPos];
            arr[randomPos] = tmp;
        }
    }

    public static int gcd(int x, int y) {
        if (y == 0)
            return x;
        return gcd(y, x % y);
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

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

        public double nextDouble() {
            return Double.valueOf(next());
        }

        String nextLine() throws IOException {
            return reader.readLine();
        }

    }
}
