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 = 6e5 + 9, N4 = N * 4, p = 1000000007;
void scan(int &x) {
  int ty = 1;
  x = 0;
  char ch = getchar();
  while (!(ch >= '0' && ch <= '9') && ch != '-') ch = getchar();
  if (ch == '-') {
    ty = -1;
    ch = getchar();
  }
  while ((ch >= '0' && ch <= '9')) {
    x *= 10;
    x += (ch - '0');
    ch = getchar();
  }
  x *= ty;
}
char s[20];
void print(int x) {
  int len = 0;
  for (; x; x /= 10) s[len++] = (x % 10) + '0';
  for (int i = len - 1; i >= 0; i--) putchar(s[i]);
}
int main() {
  int n, m, k;
  scan(n);
  scan(m);
  scan(k);
  char a = '0', b = '1';
  if (n < m) {
    swap(n, m);
    swap(a, b);
  }
  if (k % 2 == 0) {
    for (int j = 1; j <= n - k / 2; j++) putchar(a);
    for (int j = 1; j <= m - k / 2 + 1; j++) putchar(b);
    for (int j = 3; j <= k; j += 2) {
      putchar(a);
      putchar(b);
    }
    putchar(a);
  } else {
    for (int j = 1; j <= n - k / 2; j++) putchar(a);
    for (int j = 1; j <= m - k / 2; j++) putchar(b);
    for (int j = 3; j <= k; j += 2) {
      putchar(a);
      putchar(b);
    }
  }
  return 0;
}
