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;
template <typename T, typename U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
template <typename T>
inline void gn(T &first) {
  char c, sg = 0;
  while (c = getchar(), (c > '9' || c < '0') && c != '-')
    ;
  for ((c == '-' ? sg = 1, c = getchar() : 0), first = 0; c >= '0' && c <= '9';
       c = getchar())
    first = (first << 1) + (first << 3) + c - '0';
  if (sg) first = -first;
}
template <class T, class T1>
inline void gn(T &first, T1 &second) {
  gn(first);
  gn(second);
}
template <class T, class T1, class T2>
inline void gn(T &first, T1 &second, T2 &z) {
  gn(first);
  gn(second);
  gn(z);
}
template <typename T>
inline void print(T first) {
  if (first < 0) {
    putchar('-');
    return print(-first);
  }
  if (first < 10) {
    putchar('0' + first);
    return;
  }
  print(first / 10);
  putchar(first % 10 + '0');
}
template <typename T>
inline void println(T first) {
  print(first), putchar('\n');
}
template <typename T>
inline void printsp(T first) {
  print(first), putchar(' ');
}
template <class T, class T1>
inline void print(T first, T1 second) {
  printsp(first), println(second);
}
template <class T, class T1, class T2>
inline void print(T first, T1 second, T2 z) {
  printsp(first), printsp(second), println(z);
}
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
int prime[2020], flag[2020];
map<long long, int> mp;
int calc(int a, int b, int c, int n) {
  if (a > b) swap(a, b);
  if (a > c) swap(a, c);
  if (b > c) swap(b, c);
  if (n == 0) return 1;
  long long id =
      ((long long)a << 36) + ((long long)b << 24) + ((long long)c << 12) + n;
  if (mp.count(id)) return mp[id];
  int ans = 0, p = prime[n];
  for (int i = 0, first = a; first; i++, first /= p) {
    for (int j = 0, second = b; second; j++, second /= p) {
      for (int k = 0, z = c; z; k++, z /= p) {
        ans += ((long long)(i + j + k + 1) * calc(first, second, z, n - 1)) &
               ((1 << 30) - 1);
        ans &= ((1 << 30) - 1);
      }
    }
  }
  return mp[id] = ans;
}
int main() {
  int a, b, c;
  gn(a, b, c);
  int np = 0;
  prime[0] = 1;
  for (int i = 2; i < 2020; i++) {
    if (!flag[i]) prime[++np] = i, flag[i] = 1;
    for (int j = 1; j <= np && prime[j] * i < 2020; j++) {
      flag[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
  println(calc(a, b, c, np));
  return 0;
}