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 Set(long long N, long long pos) { return N = N | (1 << pos); }
long long reset(long long N, long long pos) { return N = N & ~(1 << pos); }
bool check(long long N, long long pos) { return (bool)(N & (1 << pos)); }
int arr[200];
vector<int> ok;
int main() {
  int a, b, x;
  cin >> a >> b >> x;
  string ans = "";
  int n = a + b;
  int z = 0, o = 0;
  int cnt = 0;
  char X = '0', Y = '1';
  if (a >= b) {
    z = 1;
    X = '0';
    Y = '1';
  } else {
    o = 1;
    X = '1';
    Y = '0';
  }
  while (true) {
    if (ans.size() == x) break;
    if (cnt % 2 == 0) {
      cnt++;
      ans += X;
      if (X == '1')
        b--;
      else
        a--;
      continue;
    }
    if (cnt % 2 == 1) {
      cnt++;
      ans += Y;
      if (Y == '1')
        b--;
      else
        a--;
    }
  }
  if (ans.size() == 0) {
    for (int i = 0; i < a; i++) ans += '0';
    for (int i = 0; i < b; i++) ans += '1';
  } else {
    if (ans[ans.size() - 1] == '1') {
      for (int i = 0; i < b; i++) ans += '1';
      for (int i = 0; i < a; i++) ans += '0';
    } else {
      for (int i = 0; i < a; i++) ans += '0';
      for (int i = 0; i < b; i++) ans += '1';
    }
  }
  cout << ans << endl;
  return 0;
}
