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;
long long int mul(long long int x, long long int y) {
  return ((x % int(1e9 + 7)) * (y % int(1e9 + 7))) % int(1e9 + 7);
}
long long int add(long long int x, long long int y) {
  return ((x) + (y)) % int(1e9 + 7);
}
long long int p(long long int a, long long int b) {
  if (b == 0) return 1;
  long long int res = p(a, b / 2) % int(1e9 + 7);
  if (b & 1) return mul(a, mul(res, res));
  return mul(res, res);
}
int main() {
  {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
  };
  int a, b, x;
  cin >> a >> b >> x;
  if (x == 0) {
    while (a > 0) cout << "0", a--;
    while (b > 0) cout << "1", b--;
    cout << '\n';
    return 0;
  }
  int n = a + b;
  int cnt = 1;
  a--, b--;
  string t = a > b ? "01" : "10", second;
  second = t;
  while (cnt + 2 <= x) cnt += 2, second += t, a--, b--;
  if (cnt < x) {
    if (a > 0 && b > 0) {
      if (second[int(second.size()) - 1] == '1') {
        while (b > 0) b--, second += "1";
        while (a > 0) a--, second += "0";
      } else {
        while (a > 0) a--, second += "0";
        while (b > 0) b--, second += "1";
      }
    } else {
      while (a > 0) a--, second += "0";
      while (b > 0) b--, second += "1";
    }
    cout << second << '\n';
    return 0;
  }
  if (second[int(second.size()) - 1] == '0') {
    while (a > 0) a--, second += "0";
    while (b > 0) b--, second = "1" + second;
  } else {
    while (b > 0) b--, second += "1";
    while (a > 0) a--, second = "0" + second;
  }
  cout << second << '\n';
  return 0;
}
