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. 
#include <bits/stdc++.h>
using namespace std;
int main() {
  int t = 1;
  while (t--) {
    int a, b, x;
    cin >> a >> b >> x;
    string s1;
    int ones = 0;
    int zeros = 0;
    if (a >= b) {
      if (x % 2 == 1) {
        for (int i = 0; i < (x + 1) / 2; i++) {
          s1 += '0';
          s1 += '1';
        }
      } else {
        for (int i = 0; i < (x) / 2; i++) {
          s1 += '0';
          s1 += '1';
        }
      }
    } else {
      if (x % 2 == 1) {
        for (int i = 0; i < (x + 1) / 2; i++) {
          s1 += '1';
          s1 += '0';
        }
      } else {
        for (int i = 0; i < (x) / 2; i++) {
          s1 += '1';
          s1 += '0';
        }
      }
    }
    int x1 = s1.length();
    for (int i = 0; i < x1; i++) {
      if (s1[i] == '0')
        zeros++;
      else
        ones++;
    }
    if (a >= b) {
      if (x % 2 == 1) {
        if ((a - zeros) >= 0 && (b - ones) >= 0) {
          for (int i = 0; i < a - zeros; i++) cout << '0';
          cout << s1;
          for (int i = 0; i < b - ones; i++) cout << '1';
        }
      } else {
        if ((a - zeros) >= 0 && (b - ones) >= 0) {
          for (int i = 0; i < a - zeros - 1; i++) cout << '0';
          cout << s1;
          for (int i = 0; i < b - ones; i++) cout << '1';
          cout << '0';
        }
      }
    } else {
      if (x % 2 == 1) {
        if ((a - zeros) >= 0 && (b - ones) >= 0) {
          for (int i = 0; i < b - ones; i++) cout << '1';
          cout << s1;
          for (int i = 0; i < a - zeros; i++) cout << '0';
        }
      } else {
        if ((a - zeros) >= 0 && (b - ones) >= 0) {
          for (int i = 0; i < b - ones - 1; i++) cout << '1';
          cout << s1;
          for (int i = 0; i < a - zeros; i++) cout << '0';
          cout << '1';
        }
      }
    }
  }
}
