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>
int dr[] = {2, 2, -2, -2, 1, -1, 1, -1};
int dc[] = {1, -1, 1, -1, 2, 2, -2, -2};
int dr1[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dc1[] = {1, -1, 1, 0, -1, 0, 1, -1};
int dr2[] = {0, 0, 1, -1};
int dc2[] = {1, -1, 0, 0};
using namespace std;
int main() {
  long long int a, b, x, n, i;
  while (cin >> a >> b >> x) {
    vector<long long int> v;
    long long int p = x + 1;
    if (a >= b) {
      long long int a1 = p / 2 + p % 2;
      long long int b1 = p / 2;
      for (i = 1; i <= p; i++) {
        if (i % 2 == 1)
          v.push_back(0);
        else
          v.push_back(1);
      }
      a -= a1;
      b -= b1;
    } else {
      long long int b1 = p / 2 + p % 2;
      long long int a1 = p / 2;
      for (i = 1; i <= p; i++) {
        if (i % 2 == 1)
          v.push_back(1);
        else
          v.push_back(0);
      }
      a -= a1;
      b -= b1;
    }
    if (v[0] == 0) {
      for (long long int i = 0; i < a; i++) cout << "0";
      for (long long int i = 0; i < v.size(); i++) {
        cout << v[i];
        if (i == 1) {
          for (long long int j = 0; j < b; j++) cout << "1";
        }
      }
    } else {
      for (long long int i = 0; i < b; i++) cout << "1";
      for (long long int i = 0; i < v.size(); i++) {
        cout << v[i];
        if (i == 1) {
          for (long long int j = 0; j < a; j++) cout << "0";
        }
      }
    }
    cout << endl;
  }
}
