Let's denote d(n) as the number of divisors of a positive integer n. You are given three integers a, b and c. Your task is to calculate the following sum:

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Find the sum modulo 1073741824 (230).

Input

The first line contains three space-separated integers a, b and c (1 ≤ a, b, c ≤ 100).

Output

Print a single integer — the required sum modulo 1073741824 (230).

Examples

Input

2 2 2


Output

20


Input

5 6 7


Output

1520

Note

For the first example.

  * d(1·1·1) = d(1) = 1; 
  * d(1·1·2) = d(2) = 2; 
  * d(1·2·1) = d(2) = 2; 
  * d(1·2·2) = d(4) = 3; 
  * d(2·1·1) = d(2) = 2; 
  * d(2·1·2) = d(4) = 3; 
  * d(2·2·1) = d(4) = 3; 
  * d(2·2·2) = d(8) = 4. 



So the result is 1 + 2 + 2 + 3 + 2 + 3 + 3 + 4 = 20.
#include <bits/stdc++.h>
int a, b, c;
int vol[101][101];
int main() {
  int i, j, k, l;
  unsigned int ans, tem;
  memset(vol, 0, sizeof(vol));
  for (i = 1; i <= 100; i++) {
    for (j = 2, k = i; k > 1; j++) {
      while (k % j == 0) {
        vol[i][j]++;
        k /= j;
      }
    }
  }
  ans = 0;
  scanf("%d%d%d", &a, &b, &c);
  for (i = 1; i <= a; i++) {
    for (j = 1; j <= b; j++) {
      for (k = 1; k <= c; k++) {
        tem = 1;
        for (l = 2; l < 100; l += 2) {
          tem *= (vol[i][l] + vol[j][l] + vol[k][l] + 1);
          if (l == 2) l--;
        }
        ans += tem;
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}
