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;
static const int maxn = 2000 + 5;
static int a, b, c;
static int pri[maxn], mob[maxn], gcd[maxn][maxn];
static int vis[maxn], cnt;
static long long ans;
inline int find(int a, int b) { return b ? find(b, a % b) : a; }
inline void getmob() {
  mob[1] = 1;
  for (int i = 2; i <= 2000; ++i) {
    if (!vis[i]) pri[++cnt] = i, mob[i] = -1;
    for (int j = 1; j <= cnt && i * pri[j] <= 2000; ++j) {
      vis[i * pri[j]] = 1;
      mob[i * pri[j]] = -mob[i];
      if (i % pri[j] == 0) {
        mob[i * pri[j]] = 0;
        break;
      }
    }
  }
}
inline void getgcd() {
  for (int i = 1; i <= 2000; ++i)
    for (int j = 1; j <= 2000; ++j) gcd[i][j] = find(i, j);
}
inline long long cal(int a, int x) {
  long long r = 0;
  for (int i = 1; i <= a; ++i)
    if (find(i, x) == 1) r += (long long)(a / i);
  return r;
}
int main() {
  getmob();
  getgcd();
  scanf("%d%d%d", &a, &b, &c);
  for (int i = 1; i <= a; ++i)
    for (int j = min(b, c); j; --j)
      if (find(i, j) == 1)
        ans = (ans + (long long)(a / i) * mob[j] * cal(b / j, i) *
                         cal(c / j, i) % 1073741824) %
              1073741824;
  printf("%I64d", ans);
  return 0;
}