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;
int r[1000];
char s[100005];
int main() {
  int i, j, k, l, n, x, a, b;
  cin >> a >> b >> x;
  if (x == 1) {
    string r = "";
    while (a--) r += "0";
    while (b--) r += "1";
    cout << r << endl;
    return 0;
  } else {
    int flag = 0, c1 = 0, c2 = 0;
    char ch, ch2;
    if (a > b)
      flag = 1;
    else if (b > a)
      flag = 2;
    int temp = x - 1;
    if (flag == 1) {
      ch = '0';
      ch2 = '1';
    } else {
      ch = '1';
      ch2 = '0';
    }
    for (i = 0; i <= temp; i += 2) {
      s[i] = ch;
      c1++;
    }
    for (i = 1; i <= temp; i += 2) {
      s[i] = ch2;
      c2++;
    }
    int temp1, temp2;
    if (ch == '0') {
      temp1 = a - c1;
      temp2 = b - c2;
      int q = temp + 1;
      if (s[temp] == ch) {
        while (temp1--) {
          s[q++] = ch;
        }
        while (temp2--) {
          s[q++] = ch2;
        }
      } else {
        while (temp2--) {
          s[q++] = ch2;
        }
        while (temp1--) {
          s[q++] = ch;
        }
      }
      s[q] = '\0';
    } else if (ch == '1') {
      temp1 = b - c1;
      temp2 = a - c2;
      int q = temp + 1;
      if (s[temp] == ch) {
        while (temp1--) {
          s[q++] = ch;
        }
        while (temp2--) {
          s[q++] = ch2;
        }
      } else {
        while (temp2--) {
          s[q++] = ch2;
        }
        while (temp1--) {
          s[q++] = ch;
        }
      }
      s[q] = '\0';
    }
    for (i = 0; s[i] != '\0'; i++) cout << s[i];
    cout << endl;
  }
  return 0;
}
