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 M = 4000005, p = 1 << 30;
int n, m, K;
int f[M], S[M], P[M], Q[M];
int mo[M], prim[M], len;
int sum[M];
bool vis[M];
void init() {
  sum[1] = mo[1] = 1;
  for (int i = 2; i < M; i++) {
    if (!vis[i]) prim[++len] = i, mo[i] = -1;
    for (int j = 1; j <= len; j++) {
      int t = i * prim[j];
      if (t >= M) break;
      vis[t] = 1;
      if (i % prim[j] == 0) {
        mo[t] = 0;
        break;
      } else {
        mo[t] = -mo[i];
      }
    }
    sum[i] = sum[i - 1] + mo[i];
  }
}
int main() {
  init();
  scanf("%d %d %d", &n, &m, &K);
  if (n > m) swap(n, m);
  if (m > K) swap(m, K);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      f[i * j]++;
    }
  }
  for (int i = 1; i <= n * m; i++) {
    for (int j = 1; j <= n * m / i; j++) {
      S[i] += f[j * i];
      S[i] %= p;
    }
  }
  for (int i = 1; i <= K; i++) {
    for (int j = 1; j <= n * m / i; j++) {
      P[i] += S[i * j];
      P[i] %= p;
    }
  }
  for (int i = 1; i <= K; i++) {
    for (int j = 1; j <= i; j++) {
      Q[i] += i / j;
      Q[i] %= p;
    }
  }
  int ans = 0;
  for (int i = 1; i <= K; i++) {
    ans += 1ll * mo[i] * P[i] * Q[K / i] % p;
    ans %= p;
  }
  printf("%d\n", (ans % p + p) % p);
  return 0;
}