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 Main {
    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public Scanner(FileReader fileReader) {
            br = new BufferedReader(fileReader);
        }

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

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

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

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

        public boolean ready() throws IOException {
            return br.ready();
        }
    }
    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);PrintWriter pw = new PrintWriter(System.out);
        int a=sc.nextInt(),b=sc.nextInt(),x=sc.nextInt();int st=0;
        StringBuilder sb = new StringBuilder();
        if (a>=b){
        sb.append('0');a--;
        while (x>1){
            sb.append('1');b--;x--;if (x==1){st=1;break;}
            sb.append('0');a--;x--;
        }
        if (st==0){
            while (a>0){sb.append('0');a--;}
            while (b>0){sb.append('1');b--;}
        }
        else {
            while (b>0){sb.append('1');b--;}
            while (a>0){sb.append('0');a--;}
        }
        System.out.println(sb);
    }
        else {
            sb.append('1');b--;
            while (x>1){
                sb.append('0');a--;x--;if (x==1){st=1;break;}
                sb.append('1');b--;x--;
            }
            if (st==1){
                while (a>0){sb.append('0');a--;}
                while (b>0){sb.append('1');b--;}
            }
            else {
                while (b>0){sb.append('1');b--;}
                while (a>0){sb.append('0');a--;}
            }
            System.out.println(sb);
        }
    }
}