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.*;
import java.text.*;

public class Main {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in), 32768);
    static PrintWriter out = new PrintWriter(System.out);
    static StringTokenizer t;

    static String sn() {
        while (t == null || !t.hasMoreTokens()) {
            try {
                t = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                throw new RuntimeException();
            }
        }
        return t.nextToken();
    }

    static int ni() {
        return Integer.parseInt(sn());
    }

    static long nlo() {
        return Long.parseLong(sn());
    }

    public static void main(String args[]) {
        int t = 1;//ni();
        while (t-- > 0) {
            solve();
        }
        out.close();
    }

    static long mod = 998244353;

    static long pro(long a, long b) {
        return (a % mod * b % mod) % mod;
    }

    static long add(long a, long b) {
        return (a + b) % mod;
    }

    static long maxn = 500005;

    static void solve() {
        int a = ni();
        int b = ni();
        int x = ni();
        int max = Math.max(a, b);
        int min = Math.min(a, b);
        StringBuilder sb = new StringBuilder();
        if (max == a) {
            int cur = 0;
            while (max > 0 || min > 0) {
                if (cur == 0 && max > 0) {
                    sb.append(cur);
                    max -= 1;
                }
                if (cur == 1 && min > 0) {
                    sb.append(cur);
                    min -= 1;
                }
                cur = (cur + 1) % 2;
            }
        } else {
            int cur = 1;
            while (max > 0 || min > 0) {
                if (cur == 0 && min > 0) {
                    sb.append(cur);
                    min -= 1;
                }
                if (cur == 1 && max > 0) {
                    sb.append(cur);
                    max -= 1;
                }
                cur = (cur + 1) % 2;
            }
        }
        int c1 = 0, c2 = 0, c3 = x;
        char c[] = sb.toString().toCharArray();
        for (int i = 0; i < (a + b - 1) && c3 > 0; i++) {
            if (c[i] != c[i + 1]) {
                c3 -= 1;
                if (c[i] == '0')
                    c1 += 1;
                else
                    c2 += 1;
            }
        }
        if (c1 == a || c2 == b)
            out.println(sb);
        else {
            a -= c1;
            b -= c2;
            String s = sb.toString().substring(0, x);
            c = s.toCharArray();
            int len = c.length;
            StringBuilder ans = new StringBuilder(s);
            if (c[len - 1] == '1') {
                while (b-- > 0)
                    ans.append('1');
                while (a-- > 0)
                    ans.append('0');
            } else {
                while (a-- > 0)
                    ans.append('0');
                while (b-- > 0)
                    ans.append('1');
            }
            out.println(ans);
        }
//        out.println(sb);
    }
}