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 n, m, i, j, k, l;
long long cnt = 0, ans = 0;
string s = "";
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int a, b, x;
  cin >> a >> b >> x;
  n = a + b;
  if (a >= b) {
    s += "0";
    for (i = 0; i < x; i++) {
      if (i % 2 == 0)
        s += "1";
      else
        s += "0";
    }
    l = s.length();
    for (i = 0; i < l; i++) {
      if (s[i] == '1')
        k++;
      else
        j++;
    }
    if (b != k) {
      string s1 = "";
      for (i = 0; i < abs(b - k); i++) {
        s1 += "1";
      }
      s.erase(s.begin() + 0);
      s = s1 + s;
      s = "0" + s;
    }
    if (a != j) {
      string s1 = "";
      for (i = 0; i < abs(a - j); i++) {
        s1 += "0";
      }
      s = s1 + s;
    }
    cout << s;
  } else {
    s += "1";
    for (i = 0; i < x; i++) {
      if (i % 2 == 0)
        s += "0";
      else
        s += "1";
    }
    l = s.length();
    for (i = 0; i < l; i++) {
      if (s[i] == '1')
        k++;
      else
        j++;
    }
    if (a != j) {
      string s1 = "";
      for (i = 0; i < abs(a - j); i++) {
        s1 += "0";
      }
      s.erase(s.begin() + 0);
      s = s1 + s;
      s = "1" + s;
    }
    if (b != k) {
      string s1 = "";
      for (i = 0; i < abs(b - k); i++) {
        s1 += "1";
      }
      s = s1 + s;
    }
    cout << s;
  }
  return 0;
}
