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;
int a, b, c, N;
int gcd[2005][2005];
int pr[2005], miu[2005], vis[2005], cnt;
long long ans;
void Find_prime() {
  int i, j;
  miu[1] = 1;
  for (i = 2; i <= N; i++) {
    if (!vis[i]) pr[++cnt] = i, miu[i] = -1;
    for (j = 1; j <= cnt && pr[j] * i <= N; j++) {
      vis[i * pr[j]] = 1;
      if (i % pr[j] == 0) {
        miu[i * pr[j]] = 0;
        break;
      } else
        miu[i * pr[j]] = miu[i] * miu[pr[j]];
    }
  }
}
int gd(int a, int b) {
  if (!b) return a;
  return gd(b, a % b);
}
int main() {
  int d, i, j, k;
  long long get, tmp, ggg;
  scanf("%d %d %d", &a, &b, &c);
  N = max(max(a, b), c);
  Find_prime();
  for (i = 1; i <= N; i++)
    for (j = 1; j <= N; j++) gcd[i][j] = gd(i, j);
  for (d = 1; d <= N; d++)
    if (miu[d]) {
      for (i = 1, ggg = 0; i <= a; i++)
        if (gcd[i][d] == 1) {
          for (j = 1, get = 0; j <= b / d; j++)
            if (gcd[i][j] == 1) get = (get + (b / (d * j))) % (1 << 30);
          for (k = 1, tmp = 0; k <= c / d; k++)
            if (gcd[i][k] == 1) tmp = (tmp + (c / (d * k))) % (1 << 30);
          ggg = (ggg + tmp * get % (1 << 30) * (a / i) % (1 << 30)) % (1 << 30);
        }
      ans = (ans + ggg * miu[d] % (1 << 30) + (1 << 30)) % (1 << 30);
    }
  printf("%I64d\n", ans);
  return 0;
}