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 MOD(int a, int b) {
  if (a > b)
    return a - b;
  else
    return b - a;
}
int max3(int a, int b, int c) { return max(c, max(a, b)); }
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int main() {
  int a, b, x;
  cin >> a >> b >> x;
  if (x > 1) {
    if (a < b) {
      for (int i = 1; i <= x; i++) {
        cout << i % 2;
        if (i % 2 == 0)
          a--;
        else
          b--;
      }
      if (x % 2 == 0) {
        while (a--) cout << 0;
        while (b--) cout << 1;
      } else {
        while (b > 0) {
          cout << 1;
          b--;
        }
        while (a--) cout << 0;
      }
    } else {
      for (int i = 1; i <= x; i++) {
        cout << !(i % 2);
        if (i % 2 == 0)
          b--;
        else
          a--;
      }
      if (x % 2 == 0) {
        while (b--) cout << 1;
        while (a--) cout << 0;
      } else {
        while (a > 0) {
          cout << 0;
          a--;
        }
        while (b--) cout << 1;
      }
    }
  } else {
    while (b--) cout << 1;
    while (a--) cout << 0;
  }
}
