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>
#pragma GCC optimize("Ofast")
using namespace std;
const long double eps = 1e-7;
const int inf = 1000000010;
const long long INF = 10000000000010LL;
const int mod = 1 << 30;
const int MAXN = 2000;
int n, m, u, v, x, y, t, a, b, c, ans;
bool sieve[MAXN];
int prime[MAXN], cnt;
unordered_map<long long, int> dp[305];
long long f(int a, int b, int c, int t) {
  if (t == -1) return 1;
  long long A = (2001ll * a + b) * 2001 + c;
  if (dp[t].count(A)) return dp[t][A];
  long long p = prime[t--], res = 0;
  for (int x = 0, aa = 1; aa <= a; x++, aa *= p) {
    for (int y = 0, bb = 1; bb <= b; y++, bb *= p) {
      for (int z = 0, cc = 1; cc <= c; z++, cc *= p) {
        res = (res + (x + y + z + 1) * f(a / aa, b / bb, c / cc, t)) % mod;
      }
    }
  }
  return dp[t + 1][A] = res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  for (long long i = 2; i < MAXN; i++) {
    if (sieve[i]) continue;
    prime[cnt++] = i;
    for (long long j = i * i; j < MAXN; j += i) sieve[j] = true;
  }
  cin >> a >> b >> c;
  cout << f(a, b, c, cnt - 1) << endl;
  return 0;
}