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 = 1000;
const double eps = 1e-9;
long long INF = 1e9;
const int OO = 0x3f3f3f3f;
int dx[] = {1, -1, 0, 0, 1, -1, 1, -1};
int dy[] = {0, 0, 1, -1, 1, -1, -1, 1};
void read(string str, int node) {
  stringstream ss;
  int number;
  ss << str;
  while (ss >> number) cout << number << endl;
}
long long fp(long long b, long long p) {
  if (p == 0) return 1;
  if (p % 2 == 0) return fp((b * b) % INF, p / 2);
  return (fp(b, p - 1) * b) % INF;
}
unsigned long long read_binary(string s) {
  unsigned long long x = 0, tmp = 1;
  for (int i = s.size() - 1; i >= 0; i--) {
    x = (x + tmp * (s[i] - '0')) % INF;
    tmp = (tmp * 2) % INF;
  }
  return x;
}
long long com(long long x, long long y) {
  long long sum = 1;
  for (int i = 1; i <= y; i++) {
    sum *= x;
    x--;
    sum /= i;
  }
  return sum;
}
long long gcd(long long n1, long long n2) {
  if (n2 != 0)
    return gcd(n2, n1 % n2);
  else
    return n1;
}
int setBit(int num, int idx, int val = 1) {
  return (val) ? (num | (1 << idx)) : (num & ~(1 << idx));
}
int getBit(int num, int idx) { return ((num >> idx) & 1) == 1; }
int cntBit(int num) {
  int ret = 0;
  while (num) {
    if (num % 2) ret++;
    num /= 2;
  }
  return ret;
}
bool sum[1500];
void sieve() {
  sum[0] = sum[1] = 1;
  for (int i = 2; i * i < 1002; i++) {
    if (!sum[i]) {
      for (int j = 2 * i; j < 1002; j += i) {
        sum[j] = 1;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  int a, b, x;
  cin >> a >> b >> x;
  int xa = a, xb = b;
  deque<int> ar;
  ar.push_back(0);
  a--;
  while (x > 0) {
    if (ar.back() == 0) {
      if (b > 0) {
        ar.push_back(1), x--;
        b--;
      } else
        break;
    } else if (ar.back() == 1) {
      if (a > 0) {
        ar.push_back(0), x--;
        a--;
      } else
        break;
    }
  }
  while (a > 0) ar.push_front(0), a--;
  if (b > 0) {
    if (ar.back() == 1) {
      while (b > 0) ar.push_back(1), b--;
    } else {
      ar.pop_back();
      while (b > 0) ar.push_back(1), b--;
      ar.push_back(0);
    }
  }
  if (x > 0) {
    if (ar.back() == 1) {
      ar.push_front(ar.back());
      ar.pop_back();
    } else if (ar.front() == 0) {
      ar.push_back(ar.front());
      ar.pop_front();
    }
  }
  for (int i = 0; i < ar.size(); i++) cout << ar[i];
  cout << endl;
}
