Problem: The Little Elephant is playing with the Cartesian coordinates' system. Most of all he likes playing with integer points. The Little Elephant defines an integer point as a pair of integers (x; y), such that 0 ≤ x ≤ w and 0 ≤ y ≤ h. Thus, the Little Elephant knows only (w + 1)·(h + 1) distinct integer points.

The Little Elephant wants to paint a triangle with vertexes at integer points, the triangle's area must be a positive integer. For that, he needs to find the number of groups of three points that form such triangle. At that, the order of points in a group matters, that is, the group of three points (0;0), (0;2), (2;2) isn't equal to the group (0;2), (0;0), (2;2).

Help the Little Elephant to find the number of groups of three integer points that form a nondegenerate triangle with integer area.

Input

A single line contains two integers w and h (1 ≤ w, h ≤ 4000).

Output

In a single output line print an integer — the remainder of dividing the answer to the problem by 1000000007 (109 + 7).

Examples

Input

2 1


Output

36


Input

2 2


Output

240

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int func1(int a, int b) {
  return ((a * b) << 2) - ((a + b) << 1) - (gcd(a, b) << 1);
}
inline int func2(int a, int b) { return ((a * b) << 1) - (gcd(a, b) << 1); }
inline int func3(int a, int b) {
  return 5 * a * b - ((a + b) << 1) - (gcd(a, b) << 1);
}
int call(int a, int b) {
  if ((a & 1) && !(b & 1)) return func1(a, b);
  if (!(a & 1) && (b & 1)) return func1(b, a);
  if ((a & 1) && (b & 1))
    return func2(a, b);
  else
    return func3(a, b);
}
int main() {
  long long a, b, ans = 0;
  scanf("%lld%lld", &a, &b);
  for (int i = 1; i <= a; ++i)
    for (int j = 1; j <= b; ++j) {
      long long temp, sz = (a - i + 1) * (b - j + 1);
      if ((i & 1) && (j & 1))
        temp = 0;
      else
        temp = ((i - 1) << 1) + ((j - 1) << 1) + 4;
      temp = (temp + call(i, j)) % mod;
      temp = temp * sz % mod;
      ans = (ans + temp) % mod;
    }
  ans = ans * 6 % mod;
  printf("%lld\n", ans);
}