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>
const long long INF = 1e18;
using namespace std;
long long POWER(long long base, long long expo) {
  long long num = 1;
  long long i = expo;
  while (i) {
    num = num * base;
    i--;
  }
  return num;
}
int main() {
  long long t = 1;
  while (t--) {
    long long a, b, c, sum = 0;
    cin >> a >> b >> c;
    vector<long long> dp(a * b * c + 1, 1);
    for (long long i = 2; i <= a * b * c; i++) {
      long long x = i;
      while (x <= a * b * c) {
        dp[x]++;
        x += i;
      }
    }
    for (long long i = 1; i <= a; i++) {
      for (long long j = 1; j <= b; j++) {
        for (long long k = 1; k <= c; k++) {
          sum += (dp[i * j * k]) % (1073741824);
        }
      }
    }
    cout << sum;
  }
  return 0;
}
