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 MOD = 1000000007;
const long long LINF = 4611686018427387903;
const int dir4[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
const int dir8[8][2] = {{1, 0}, {0, 1},   {-1, 0}, {0, -1},
                        {1, 1}, {-1, -1}, {1, -1}, {-1, 1}};
const int dir63[6][3] = {{0, 0, 1},  {0, 0, -1}, {0, 1, 0},
                         {0, -1, 0}, {1, 0, 0},  {-1, 0, 0}};
const int daym[2][13] = {{0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},
                         {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}};
int GCD(int a, int b) { return b ? GCD(b, a % b) : a; }
char s[211];
int main() {
  int a, b, n;
  scanf("%d %d %d", &a, &b, &n);
  if (a < b) {
    s[0] = '1';
    b--;
  } else {
    s[0] = '0';
    a--;
  }
  int p = 1;
  while (n > 1) {
    n--;
    if (s[p - 1] == '1') {
      s[p++] = '0';
      a--;
    } else {
      s[p++] = '1';
      b--;
    }
  }
  if (s[p - 1] == '0') {
    while (a--) s[p++] = '0';
    while (b--) s[p++] = '1';
    s[p] = '\0';
  } else {
    while (b--) s[p++] = '1';
    while (a--) s[p++] = '0';
    s[p] = '\0';
  }
  printf("%s\n", s);
  return 0;
}
