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;
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
template <typename P>
P dectobin(P a) {
  if (a == 0)
    return 0;
  else
    return (a % 2 + 10 * dectobin(a / 2));
}
template <typename Y>
Y bintodec(Y a) {
  long long ans = 0, b = 1, t = a;
  while (t) {
    long long ld = t % 10;
    t /= 10;
    ans += ld * b;
    b = b * 2;
  }
  return ans;
}
template <typename H>
H modExp(H x, H n) {
  long long r = 1;
  while (n > 0) {
    if (n % 2 == 1) {
      r = (r * x) % 1000000007;
    }
    x = (x * x) % 1000000007;
    n /= 2;
  }
  return r;
}
template <typename T>
T isPowerOfTwo(T x) {
  return x && (!(x & (x - 1)));
}
template <typename T>
T lcm(T a, T b) {
  return a * b / gcd(a, b);
}
long long v[1000001];
long long arr[1000001];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ostream_iterator<long long> output(cout, " ");
  memset(arr, 0, sizeof(arr));
  for (long long i = 1000000; i >= 1; i--) {
    for (long long j = i; j <= 1000000; j += i) v[j]++;
  }
  long long a, b, c, ans = 0;
  cin >> a >> b >> c;
  for (long long(i) = (1); i < (a + 1); i++) {
    for (long long(j) = (1); j < (b + 1); j++) {
      for (long long(k) = (1); k < (c + 1); k++) {
        if (arr[i * j * k] == 0) {
          arr[i * j * k] = v[i * j * k];
        }
        ans += arr[i * j * k];
        ans %= 1073741824;
      }
    }
  }
  cout << ans;
  return 0;
}
