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 ≤ 100).

Output

Print a single integer — the required sum modulo 1073741824 (230).

Examples

Input

2 2 2


Output

20


Input

5 6 7


Output

1520

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.
#include <bits/stdc++.h>
using namespace std;
const int maxN = 1e6, maxS = 1001, M = 1073741824;
int sieve[maxS + 1];
vector<int> primes;
long long d[maxN + 1];
void calcPrime() {
  sieve[1] = 1;
  for (int x = 2; x <= maxS; ++x) {
    if (sieve[x]) continue;
    for (int u = x * 2; u <= maxS; u += x) {
      sieve[u] = x;
    }
  }
}
long long calcD(int i) {
  if (d[i] != 0) return d[i];
  int n = i;
  long long ans = 1;
  for (int x = 2; x <= i && x <= maxS; ++x) {
    if (sieve[x] != 0) continue;
    int count = 0;
    while (n % x == 0) {
      n = n / x;
      count++;
    }
    ans *= count + 1;
  }
  d[i] = ans;
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  calcPrime();
  int a, b, c;
  cin >> a >> b >> c;
  int ans = 0;
  for (int i = 1; i <= a; ++i) {
    for (int j = 1; j <= b; ++j) {
      for (int k = 1; k <= c; ++k) {
        ans = (ans + calcD(i * j * k)) % M;
      }
    }
  }
  cout << ans;
}
