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 int N = 100005;
long long mod = 1000000007;
long long mod2 = 1000000009;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
bool is_vowel(char c) {
  if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') return 1;
  return 0;
}
double getDistance(double x1, double y1, double x2, double y2) {
  return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
long long extended_euclidean(long long a, long long b, long long &x,
                             long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long g = extended_euclidean(b, a % b, y, x);
  y -= (a / b) * x;
  return g;
}
long long power(long long base, long long p) {
  if (p == 1) return base;
  if (!p) return 1ll;
  long long ret = power(base, p / 2);
  ret *= ret;
  ret %= mod;
  if (p & 1) ret *= base;
  return ret % mod;
}
pair<long long, long long> vec(pair<long long, long long> a,
                               pair<long long, long long> b) {
  pair<long long, long long> ret =
      pair<long long, long long>(a.first - b.first, a.second - b.second);
  return ret;
}
long long getArea(pair<long long, long long> a, pair<long long, long long> b,
                  pair<long long, long long> c) {
  pair<long long, long long> x = vec(a, b);
  pair<long long, long long> y = vec(c, b);
  return abs(x.first * y.second - x.second * y.first);
}
bool intersect(const complex<double> &a, const complex<double> &b,
               const complex<double> &p, const complex<double> &q,
               complex<double> &ret) {
  double d1 = ((conj(p - a) * (b - a)).imag());
  double d2 = ((conj(q - a) * (b - a)).imag());
  ret = (d1 * q - d2 * p) / (d1 - d2);
  return fabs(d1 - d2) > 1e-9;
}
double getDistance(complex<double> a, complex<double> b) {
  return sqrt((a.real() - b.real()) * (a.real() - b.real()) +
              (a.imag() - b.imag()) * (a.imag() - b.imag()));
}
int main() {
  int a, b, x;
  cin >> a >> b >> x;
  vector<int> v;
  if (a > b)
    v.push_back(0), a--;
  else
    v.push_back(1), b--;
  while (x) {
    if (v.back() == 1)
      v.push_back(0), a--;
    else
      v.push_back(1), b--;
    x--;
  }
  for (int i = 0; i < v.size(); i++) {
    cout << v[i];
    if (v[i] == 1) {
      while (b > 0) cout << 1, b--;
    }
    if (v[i] == 0) {
      while (a > 0) cout << 0, a--;
    }
  }
  return 0;
}
