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;
template <typename T>
T gcd(T x, T y) {
  if (y == 0)
    return x;
  else
    return gcd(y, x % y);
}
const double INF = 1e+9;
const double INFLL = 1e+18;
const double EPS = 1e-9;
const double PI = acos(-1);
const int knightDir[8][2] = {{-2, -1}, {-2, 1}, {-1, 2},  {1, 2},
                             {2, -1},  {2, 1},  {-1, -2}, {1, -2}};
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, 1, 0, -1};
int main() {
  int zero, one, mix;
  cin >> zero >> one >> mix;
  string ans, temp2;
  int flag = zero > one ? 0 : 1;
  for (int i = 0; i <= mix; i++) {
    if (flag)
      temp2 += '1', one--;
    else
      temp2 += '0', zero--;
    flag = 1 - flag;
  }
  string temp;
  int l = temp2.length();
  while (one--) temp += '1';
  while (zero--) temp += '0';
  if (temp2[0] == '1') {
    ans += '1';
    ans += temp;
    for (int i = 1; i < l; i++) ans += temp2[i];
  } else {
    ans += '0';
    reverse(temp.begin(), temp.end());
    ans += temp;
    for (int i = 1; i < l; i++) ans += temp2[i];
  }
  cout << ans << endl;
}
