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 int N = 4500000, mod = 1000000007;
long long c[N][4], f[4];
template <class T>
inline T gcd(T a, T b) {
  if (a < b) swap(a, b);
  for (T r; b; r = a % b, a = b, b = r)
    ;
  return a;
}
int main() {
  int w, h;
  scanf("%d%d", &w, &h);
  c[0][0] = 1;
  for (int i = 1; i < N; i++) {
    c[i][0] = 1;
    for (int j = 1; j < 4; j++) c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
  }
  memset(f, 0, sizeof(f));
  for (int i = 0; i <= w; i++) {
    for (int j = 0; j <= h; j++) {
      f[i % 2 * 2 + j % 2]++;
    }
  }
  long long cnt = 0;
  for (int i = 1; i <= w; i++) {
    for (int j = 1; j <= h; j++) {
      int k = gcd(i, j);
      cnt += (long long)(w - i + 1) * (h - j + 1) * (k - 1) % mod;
      cnt %= mod;
    }
  }
  cnt = (cnt * 2 + c[w + 1][3] * (h + 1) % mod + c[h + 1][3] * (w + 1) % mod) %
        mod;
  long long ans = 0, tot = (w + 1) * (h + 1);
  for (int i = 0; i < 4; i++) {
    ans += c[f[i]][3];
    ans += c[f[i]][2] * (tot - f[i]) % mod;
    ans %= mod;
  }
  ans = ((ans - cnt) % mod + mod) % mod;
  cout << ans * 6 % mod << endl;
  return 0;
}