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;
inline int D() {
  int m;
  cin >> m;
  return m;
}
inline long long lD() {
  long long m;
  cin >> m;
  return m;
}
inline double dD() {
  double m;
  cin >> m;
  return m;
}
void fast() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int main() {
  fast();
  int a = D(), b = D(), x = D();
  string ans = "";
  char d;
  if (a > b) {
    ans += '0';
    a--;
    d = '0';
  } else {
    ans += '1';
    b--;
    d = '1';
  }
  x--;
  while (x--) {
    if (d == '0')
      d = '1';
    else
      d = '0';
    ans += d;
    if (d == '1')
      b--;
    else
      a--;
  }
  if (d == '1') {
    while (b--) ans += "1";
    while (a--) ans += "0";
  } else {
    while (a--) ans += "0";
    while (b--) ans += "1";
  }
  cout << ans;
}
