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 MainClass {
    public static void main(String[] args) {
        Reader in = new Reader(System.in);
        int t = 1;
        StringBuilder stringBuilder = new StringBuilder();
        while (t-- > 0) {
            int a = in.nextInt(), b = in.nextInt(), x = in.nextInt();
            if (a >= b) {
                for (int i = 0; i < x / 2; i++) {
                    stringBuilder.append("01");
                    a --;
                    b --;
                }
                if (x % 2 == 0) {
                    for (int i = 0; i < b; i++)
                        stringBuilder.append('1');
                }
                for (int i = 0; i < a; i++) {
                    stringBuilder.append("0");
                }
                if (x % 2 == 1) {
                    for (int i = 0; i < b; i++)
                        stringBuilder.append('1');
                }
            } else {
                for (int i = 0; i < x / 2; i++) {
                    stringBuilder.append("10");
                    a --;
                    b --;
                }
                if (x % 2 == 0) {
                    for (int i = 0; i < a; i++)
                        stringBuilder.append('0');
                }
                for (int i = 0; i < b; i++) {
                    stringBuilder.append("1");
                }
                if (x % 2 == 1) {
                    for (int i = 0; i < a; i++)
                        stringBuilder.append('0');
                }
            }

        }
        System.out.println(stringBuilder);
    }
}
class Reader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public Reader(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.parseLong(next());
    }

}