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 maxn = 3e5 + 10;
int a_min(int a, int b) {
  if (a < b)
    return a;
  else
    return b;
}
int a_max(int a, int b) {
  if (a > b)
    return a;
  else
    return b;
}
char s[maxn];
int main() {
  int a, b, x;
  cin >> a >> b >> x;
  int w = a_min(a, b);
  if (a > b) {
    int c0 = 0, c1 = 0;
    int tail = 0;
    printf("0");
    c0++;
    for (int i = 1; i <= x - 1; i++) {
      if (i % 2 == 1)
        printf("1"), c1++, tail = 1;
      else
        printf("0"), c0++, tail = 0;
    }
    if (tail == 0 || x - 1 == 0) {
      int temp = a - c0;
      while (temp--) printf("0");
      temp = b - c1;
      while (temp--) printf("1");
    } else {
      int temp = b - c1;
      while (temp--) printf("1");
      temp = a - c0;
      while (temp--) printf("0");
    }
    printf("\n");
  } else {
    int c11 = 0, c1 = 0;
    printf("1");
    c1++;
    int tail = 1;
    int flag = 10;
    for (int i = 1; i <= x - 1; i++) {
      if (i % 2 == 1)
        printf("0"), c11++, tail = 0;
      else
        printf("1"), c1++, tail = 1;
    }
    if (tail == 0 && x - 1 != 0) {
      int temp = a - c11;
      while (temp > 0) printf("0"), temp--;
      temp = b - c1;
      while (temp > 0) printf("1"), temp--;
    } else {
      int llll = b - c1;
      if (llll) flag++;
      while (llll > 0) printf("1"), llll--;
      llll = a - c11;
      while (llll > 0) printf("0"), llll--;
    }
    printf("\n");
  }
  return 0;
}
