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;
  cin >> a >> b >> x;
  string s = "";
  if (x & 1) {
    int add = 0;
    for (int i = 1; i <= x + 1; i++) {
      s += char(add + 48);
      add = 1 - add;
    }
    int n0, n1;
    n0 = n1 = (x + 1) / 2;
    a -= n0;
    b -= n1;
    for (int i = 1; i <= a; i++) {
      s = '0' + s;
    }
    for (int i = 1; i <= b; i++) {
      s.push_back('1');
    }
  } else {
    int add;
    int n0, n1;
    if (a > b) {
      add = 0;
      n0 = x / 2 + 1;
      n1 = x / 2;
    } else {
      add = 1;
      n0 = x / 2;
      n1 = x / 2 + 1;
    }
    for (int i = 1; i <= x + 1; i++) {
      s += char(add + 48);
      add = 1 - add;
    }
    if (a > b) {
      a -= n0;
      b -= n1;
      for (int i = 1; i <= b; i++) {
        s.insert(1, "1");
      }
      for (int i = 1; i <= a; i++) s = '0' + s;
    } else {
      a -= n0;
      b -= n1;
      for (int i = 1; i <= a; i++) {
        s.insert(1, "0");
      }
      for (int i = 1; i <= b; i++) s = '1' + s;
    }
  }
  cout << s << endl;
}
