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 a, b, x;
  string s = "";
  cin >> a >> b >> x;
  if (x == 1) int n = a + b;
  int c1 = 0;
  int c0 = 0;
  char initial;
  if (b > a) {
    initial = '1';
    c1++;
  } else {
    initial = '0';
    c0++;
  }
  s += initial;
  int k = 0;
  if (initial == '1') {
    while (k < x) {
      if (k % 2 == 0) {
        s += '0';
        c0++;
      } else {
        s += '1';
        c1++;
      }
      k++;
    }
  } else {
    k = 0;
    while (k < x) {
      if (k % 2 == 0) {
        s += '1';
        c1++;
      } else {
        s += '0';
        c0++;
      }
      k++;
    }
  }
  int m;
  string str = "";
  if (c0 != a) {
    m = 0;
    while (s[m] != '0') m++;
    k = 0;
    while (k <= m) {
      str += s[k];
      k++;
    }
    while (c0 != a) {
      str += '0';
      c0++;
    }
    k = m + 1;
    while (k < s.size()) {
      str += s[k];
      k++;
    }
    s = str;
    str = "";
  }
  if (c1 != b) {
    m = 0;
    while (s[m] != '1') {
      m++;
    }
    k = 0;
    while (k <= m) {
      str += s[k];
      k++;
    }
    while (c1 != b) {
      str += '1';
      c1++;
    }
    k = m + 1;
    while (k < s.size()) {
      str += s[k];
      k++;
    }
    s = str;
  }
  cout << s;
}
