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 T1, typename T2>
istream &operator>>(istream &is, vector<pair<T1, T2>> &v) {
  for (pair<T1, T2> &t : v) is >> t.first >> t.second;
  return is;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (T &t : v) is >> t;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (const T &t : v) {
    os << t << " ";
  }
  os << '\n';
  return os;
}
double pi = acos(-1.0);
long long md = 1000000007;
long long abst(long long a) { return a < 0 ? -a : a; }
struct HASH {
  size_t operator()(const pair<long long, long long> &x) const {
    return (size_t)x.first * 37U + (size_t)x.second;
  }
};
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
long long Pow(long long n, long long x) {
  long long ans = 1;
  while (x > 0) {
    if (x & 1) ans = (ans * n) % md;
    n = (n * n) % md;
    x = x >> 1;
  }
  return ans;
}
vector<long long> fact, inv;
void inverse(long long n) {
  inv.resize(n + 1);
  inv[0] = 1;
  for (long long i = 1; i <= n; i++) inv[i] = Pow(fact[i], md - 2);
}
void factorial(long long n) {
  fact.resize(n + 1);
  fact[0] = 1;
  for (long long i = 1; i <= n; i++) fact[i] = (fact[i - 1] * i) % md;
}
long long max(long long a, long long b) { return a > b ? a : b; }
long long min(long long a, long long b) { return a < b ? a : b; }
void solve() {
  long long a, b, x;
  cin >> a >> b >> x;
  string s;
  x--;
  if (a > b) {
    s = "0";
    a--;
    for (long long i = 0; i < x; i++) {
      if (i & 1) {
        s += "0";
        a--;
      } else {
        s += "1";
        b--;
      }
    }
    if (s.back() == '0') {
      while (a--) s += "0";
      while (b--) s += "1";
    } else {
      while (b--) s += "1";
      while (a--) s += "0";
    }
  } else {
    s = "1";
    b--;
    for (long long i = 0; i < x; i++)
      if (i & 1) {
        s += "1";
        b--;
      } else {
        a--;
        s += "0";
      }
    if (s.back() == '0') {
      while (a--) s += "0";
      while (b--) s += "1";
    } else {
      while (b--) s += "1";
      while (a--) s += "0";
    }
  }
  cout << s;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  t = 1;
  while (t--) {
    solve();
  }
}
