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>
const int Max = 200006;
const int Inf = 2e9;
const long long MOD = 998244353;
using namespace std;
string ans(int start, int a, int b, int x) {
  string s = "";
  for (int i = start; i <= start + x - 1; i++) {
    if (i & 1) {
      s += '1';
      b--;
    } else {
      s += '0';
      a--;
    }
  }
  char c;
  int ax;
  if (s[s.size() - 1] == '0')
    c = '0', ax = a;
  else
    c = '1', ax = b;
  for (int i = 1; i <= ax; i++) {
    s += c;
    if (c == '1')
      b--;
    else
      a--;
  }
  for (int i = 1; i <= a; i++) s += '0';
  for (int i = 1; i <= b; i++) s += '1';
  return s;
}
void solve() {
  int a, b, x;
  cin >> a >> b >> x;
  string ans1 = ans(0, a, b, x), ans2 = ans(1, a, b, x);
  int ok1 = 0, ok2 = 0;
  for (int i = 0; i < ans1.size() - 1; i++) {
    if (ans1[i] != ans1[i + 1]) ok1++;
  }
  for (int i = 0; i < ans2.size() - 1; i++) {
    if (ans2[i] != ans2[i + 1]) ok2++;
  }
  if (ok1 == x)
    cout << ans1 << "\n";
  else
    cout << ans2 << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
