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;
bool sortinrev(const pair<long long int, long long int> &a,
               const pair<long long int, long long int> &b) {
  if (a.first != b.first)
    return (a.first > b.first);
  else
    return (a.second < b.second);
}
long long int power(long long int x, long long int n, long long int m) {
  long long int result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % m;
    x = (x * x) % m;
    n = n / 2;
  }
  return result % m;
}
bool isPrime(long long int k) {
  if (k <= 1) {
    return false;
  }
  if (k == 2) {
    return true;
  }
  if (k % 2 == 0) {
    return false;
  }
  long long int i = 3;
  while (i * i <= k) {
    if (k % i == 0) {
      return false;
    }
    i += 2;
  }
  return true;
}
long long int bin(long long int n) {
  int di = 0;
  while (n) {
    di = di + n % 2;
    n = n / 2;
  }
  return di;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  t = 1;
  while (t--) {
    long long int a, b, x;
    cin >> b >> a >> x;
    if (a > b) {
      x--;
      cout << "1";
      a--;
      for (long long int i = 0; i < x; i++) {
        if (i % 2 == 0) {
          cout << "0";
          b--;
        } else {
          cout << "1";
          a--;
        }
      }
      if (x % 2 == 0) {
        for (long long int i = 0; i < a; i++) cout << "1";
        for (long long int i = 0; i < b; i++) cout << "0";
      } else {
        for (long long int i = 0; i < b; i++) cout << "0";
        for (long long int j = 0; j < a; j++) cout << "1";
      }
    } else {
      x--;
      cout << "0";
      b--;
      for (long long int i = 0; i < x; i++) {
        if (i % 2 == 0) {
          cout << "1";
          a--;
        } else {
          cout << "0";
          b--;
        }
      }
      if (x % 2 != 0) {
        for (long long int i = 0; i < a; i++) cout << "1";
        for (long long int i = 0; i < b; i++) cout << "0";
      } else {
        for (long long int i = 0; i < b; i++) cout << "0";
        for (long long int j = 0; j < a; j++) cout << "1";
      }
    }
  }
  return 0;
}
