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 mod = (1 << 30);
int isprime[2005], prime[2005], mobius[2005], gcd[2005][2005];
int get_gcd(int x, int y) {
  if (!y) {
    return x;
  }
  if (gcd[x][y]) {
    return gcd[x][y];
  }
  return gcd[x][y] = get_gcd(y, x % y);
}
int get(int n, int m) {
  int answer = 0;
  for (int i = 1; i <= n; i++) {
    if (get_gcd(m, i) == 1) {
      answer += n / i;
    }
  }
  return answer;
}
int main() {
  int n, m, o, now;
  int64_t answer;
  now = 0;
  mobius[1] = 1;
  for (int i = 2; i <= 2000; i++) {
    if (!isprime[i]) {
      prime[now++] = i;
      mobius[i] = -1;
    }
    for (int j = 0; j < now && i * prime[j] <= 2000; j++) {
      isprime[prime[j] * i] = 1;
      if (i % prime[j]) {
        mobius[i * prime[j]] = -mobius[i];
      } else {
        mobius[i * prime[j]] = 0;
        break;
      }
    }
  }
  while (~scanf("%d%d%d", &n, &m, &o)) {
    answer = 0;
    if (n > m) {
      swap(n, m);
    }
    if (m > o) {
      swap(m, o);
    }
    if (n > m) {
      swap(n, m);
    }
    if (m > o) {
      swap(m, o);
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        if (get_gcd(i, j) == 1) {
          answer +=
              1ll * mobius[j] * (n / i) * get(m / j, i) * get(o / j, i) % mod;
          answer %= mod;
        }
      }
    }
    printf("%I64d\n", (answer + mod) % mod);
  }
  return 0;
}