Problem: 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 ≤ 2000).

Output

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

Examples

Input

2 2 2


Output

20


Input

4 4 4


Output

328


Input

10 10 10


Output

11536

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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long d[102][102][102], q[1000005], flag[1000005], prime[1000005];
void _prime() {
  memset(q, 0, sizeof(q));
  memset(prime, 0, sizeof(prime));
  int i, j;
  for (i = 2; i < 1000005; i++) {
    if (q[i] == 0) {
      q[i] = 1;
      prime[i] = 1;
      for (j = i + i; j < 1000005; j += i) q[j]++;
    } else {
      for (j = i + i; j < 1000005; j += i) q[j]++;
    }
  }
  for (i = 2; i < 1000005; i++) {
    if (prime[i] == 1)
      q[i] = 2;
    else
      q[i] += 2;
  }
}
int main() {
  int i, j, k, a, b, c;
  _prime();
  q[1] = 1;
  for (i = 1; i <= 100; i++) {
    for (j = 1; j <= 100; j++) {
      for (k = 1; k <= 100; k++) {
        d[i][k][j] = q[i * k * j];
      }
    }
  }
  while (scanf("%d%d%d", &a, &b, &c) == 3) {
    long long cnt = 0;
    for (i = 1; i <= a; i++) {
      for (j = 1; j <= b; j++) {
        for (k = 1; k <= c; k++) {
          cnt += d[i][j][k];
        }
      }
    }
    printf("%I64d\n", cnt);
  }
  return 0;
}