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;
const int Maxn = 2050, MOD = 1073741824;
int mu[Maxn], prime[Maxn], tot, g[Maxn][Maxn];
bool vis[Maxn];
void Getmu() {
  mu[1] = 1;
  for (int i = 2; i < Maxn; ++i) {
    if (!vis[i]) {
      prime[++tot] = i;
      mu[i] = -1;
    }
    for (int j = 1; i * prime[j] < Maxn && j <= tot; ++j) {
      vis[i * prime[j]] = 1;
      if (i % prime[j] == 0) {
        mu[i * prime[j]] = 0;
        break;
      }
      mu[i * prime[j]] = -mu[i];
    }
  }
}
int gcd(int x, int y) { return x == 0 ? y : gcd(y % x, x); }
int main() {
  int a, b, c;
  scanf("%d%d%d", &a, &b, &c);
  Getmu();
  for (int i = 1; i < Maxn; ++i)
    for (int j = 1; j <= i; ++j) g[i][j] = g[j][i] = gcd(i, j);
  long long ans = 0;
  for (int i = 1; i <= a; ++i) {
    for (int d = 1; d <= min(b, c); ++d)
      if (g[i][d] == 1) {
        long long res1 = 0, res2 = 0;
        for (int j = d; j <= b; j += d)
          if (g[i][j] == 1) {
            res1 += b / j;
          }
        for (int k = d; k <= c; k += d)
          if (g[i][k] == 1) {
            res2 += c / k;
          }
        ans = (ans + 1ll * mu[d] * (a / i) * res1 * res2 % MOD) % MOD;
      }
  }
  printf("%lld\n", (ans + MOD) % MOD);
  return 0;
}