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 arr[2], x;
  cin >> arr[0] >> arr[1] >> x;
  vector<int> v;
  int curr = 1;
  if (x % 2 == 1) {
    v.push_back(curr);
    arr[curr]--;
    curr ^= 1;
    for (int i = 0; i < x; i++) {
      v.push_back(curr);
      arr[curr]--;
      curr ^= 1;
    }
    if (arr[1] > 0) {
      while (arr[1]--) {
        cout << "1";
      }
    }
    for (int i = 0; i < v.size(); i++) cout << v[i];
    if (arr[0] > 0) {
      while (arr[0]--) cout << "0";
    }
  } else {
    if (arr[0] <= arr[1]) {
      v.push_back(curr);
      arr[curr]--;
      curr ^= 1;
      for (int i = 0; i < x; i++) {
        v.push_back(curr);
        arr[curr]--;
        curr ^= 1;
      }
      if (arr[1] > 0) {
        while (arr[1]--) cout << "1";
      }
      for (int i = 0; i < v.size() - 1; i++) cout << v[i];
      if (arr[0] > 0) {
        while (arr[0]--) cout << "0";
      }
      cout << "1";
    } else {
      curr = 0;
      v.push_back(curr);
      arr[curr]--;
      curr ^= 1;
      for (int i = 0; i < x; i++) {
        v.push_back(curr);
        arr[curr]--;
        curr ^= 1;
      }
      if (arr[0] > 0) {
        while (arr[0]--) cout << "0";
      }
      for (int i = 0; i < v.size() - 1; i++) cout << v[i];
      if (arr[1] > 0) {
        while (arr[1]--) cout << "1";
      }
      cout << "0";
    }
  }
  cout << endl;
}
