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;
long long int dp[1000001];
long long int solve(long long int num) {
  if (dp[num] != -1) {
    return dp[num];
  }
  long long int x = sqrt(num);
  long long int t = 0;
  if (x * x == num) {
    for (int i = 1; i < x; i++) {
      if (num % i == 0) {
        t++;
      }
    }
    dp[num] = 2 * t + 1;
  } else {
    for (int i = 1; i <= x; i++) {
      if (num % i == 0) {
        t++;
      }
    }
    dp[num] = 2 * t;
  }
  return dp[num] % 1073741824;
}
int main() {
  long long int a, b, c;
  memset(dp, -1, sizeof(dp));
  cin >> a >> b >> c;
  long long int ans = 0;
  for (int i = 1; i <= a; i++) {
    for (int j = 1; j <= b; j++) {
      for (int k = 1; k <= c; k++) {
        long long int tmp = i * j * k;
        long long int x = solve(tmp);
        ans = (ans % 1073741824 + x % 1073741824) % 1073741824;
      }
    }
  }
  cout << ans;
}
