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;
const long long MOD = 1000 * 1000 * 1000 + 7;
const long long INF = 1ll * 1000 * 1000 * 1000 * 1000 * 1000 * 1000 + 7;
const long long MOD2 = 998244353;
const long long N = 1000 * 100 + 10;
const long long N2 = 70;
const long double PI = 3.141592653589793;
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long power(long long x, long long y, long long p = LLONG_MAX) {
  long long res = 1;
  x %= p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
signed main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  ;
  long long n0, n1, x;
  cin >> n0 >> n1 >> x;
  if (x % 2 == 0) {
    string ans = "";
    if (n0 > n1) {
      ans += '0';
      n0--;
      for (long long i = 0; i < (x / 2); i++) {
        ans += "10";
        n0--;
        n1--;
      }
      ans += (string((n0), '0'));
      bool ok = 1;
      for (long long i = 0; i < ans.length(); i++) {
        cout << ans[i];
        if (ans[i] == '1' and ok == 1) {
          ok = 0;
          while (n1--) {
            cout << '1';
          }
        }
      }
    } else {
      ans += '1';
      n1--;
      for (long long i = 0; i < (x / 2); i++) {
        ans += "01";
        n0--;
        n1--;
      }
      ans += (string((n1), '1'));
      bool ok = 1;
      for (long long i = 0; i < ans.length(); i++) {
        cout << ans[i];
        if (ans[i] == '0' and ok == 1) {
          ok = 0;
          while (n0--) {
            cout << '0';
          }
        }
      }
    }
  } else {
    string ans = "";
    for (long long i = 0; i < (long long)ceil((long double)x / 2.00); i++) {
      ans += "10";
      n1--;
      n0--;
    }
    ans += (string((n0), '0'));
    while (n1--) {
      cout << '1';
    }
    for (long long i = 0; i < ans.length(); i++) {
      cout << ans[i];
    }
  }
  return 0;
}
