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 long long MSK = (1 << 30) - 1;
const int PI = 303;
vector<int> ps;
int prime[2001];
map<long long, long long> dp;
long long to(long long a, long long b, long long c, long long i) {
  return (a << 45) | (b << 30) | (c << 15) | i;
}
long long solve(int a, int b, int c, int i) {
  int p = ps[i];
  if (a < p && b < p && c < p) return 1;
  if (a < p) a = 1;
  if (b < p) b = 1;
  if (c < p) c = 1;
  if (a > b) swap(a, b);
  if (a > c) swap(a, c);
  if (b > c) swap(b, c);
  long long X = to(a, b, c, i);
  int y = dp[X];
  if (y != 0) return y;
  long long ans = 0;
  int qa = a, qb, qc, ea = 0, eb, ec;
  while (qa > 0) {
    qb = b, eb = 0;
    while (qb > 0) {
      qc = c, ec = 0;
      while (qc > 0) {
        ans += (long long)(ea + eb + ec + 1) * solve(qa, qb, qc, i + 1);
        ans &= MSK;
        qc /= p;
        ec++;
      }
      qb /= p;
      eb++;
    }
    qa /= p;
    ea++;
  }
  return dp[X] = ans;
}
void sieve() {
  for (int p = 2; p <= 2000; p++) {
    if (prime[p]) continue;
    ps.push_back(p);
    for (int q = p; q <= 2000; q += p) prime[q] = p;
  }
}
int main() {
  sieve();
  ps.push_back(2500);
  int a, b, c;
  cin >> a >> b >> c;
  long long x = solve(a, b, c, 0);
  cout << x << '\n';
}