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:

<image>

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;
const int S = 2001;
const int P = 1073741824;
int Gcd[S][S], flag[S], prime[S], mul[S];
int ans, A, B, C, i, j, a, d, t, cnt;
int gcd(int a, int b) { return !b ? a : gcd(b, a % b); }
int F(int n, int g) {
  int ret = 0;
  for (int i = 1; i <= n; i++)
    if (Gcd[i][g] == 1) ret += n / i;
  return ret;
}
int main() {
  scanf("%d%d%d", &A, &B, &C);
  for (mul[1] = 1, i = 2; i < S; i++) {
    if (!flag[i]) prime[++cnt] = i, mul[i] = -1;
    for (j = 1; j <= cnt; j++) {
      if ((t = prime[j] * i) >= S) break;
      flag[t] = 1;
      mul[t] = -mul[i];
      if (i % prime[j] == 0) {
        mul[t] = 0;
        break;
      }
    }
  }
  for (i = 1; i < S; i++)
    for (j = 1; j < S; j++) Gcd[i][j] = gcd(i, j);
  for (a = 1; a <= A; a++) {
    for (d = 1; d <= B && d <= C; d++)
      if (Gcd[a][d] == 1) ans += mul[d] * (A / a) * F(B / d, a) * F(C / d, a);
    ans %= P;
    if (ans < 0) ans += P;
  }
  printf("%d\n", ans);
}