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 N = 2010;
const int M = 4000000;
int a, b, c, n;
int mu[M + 10], f[M + 10];
int prime[M], cnt;
bool flg[M + 10];
int fs[M + 10], cs[M + 10], gs[M + 10];
void getPrime() {
  mu[1] = 1;
  for (int i = 2; i <= M; i++) {
    if (!flg[i]) prime[++cnt] = i, mu[i] = -1;
    for (int j = 1; j <= cnt && i * prime[j] <= M; j++) {
      flg[i * prime[j]] = 1;
      if (i % prime[j] == 0) {
        mu[i * prime[j]] = 0;
        break;
      }
      mu[i * prime[j]] = -mu[i];
    }
  }
}
int main() {
  scanf("%d%d%d", &a, &b, &c);
  n = a * b;
  getPrime();
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= b; j++) f[i * j]++;
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j += i) fs[i] += f[j];
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j += i) cs[i] += c / j;
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j += i) gs[j] += mu[i] * cs[i];
  int res = 0;
  for (int i = 1; i <= n; i++) res += fs[i] * gs[i];
  printf("%d\n", res & (1 << 30) - 1);
  return 0;
}