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>
const int N = 2000;
const int mod = 1e9 + 7;
int n, m, ans, cx[2], cy[2];
int abs(int x) { return x > 0 ? x : -x; }
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int main() {
  scanf("%d %d", &n, &m);
  cx[1] = (n >> 1) + (n & 1);
  cx[0] = n + 1 - cx[1];
  cy[1] = (m >> 1) + (m & 1);
  cy[0] = m + 1 - cy[1];
  for (int x = 0; x < 2; x++) {
    for (int y = 0; y < 2; y++) {
      for (int x1 = 0; x1 < 2; x1++) {
        for (int y1 = 0; y1 < 2; y1++) {
          for (int x2 = 0; x2 < 2; x2++) {
            for (int y2 = 0; y2 < 2; y2++) {
              int S = abs((x1 - x) * (y2 - y) - (x2 - x) * (y1 - y));
              if (!(S & 1))
                ans = (1ll * ans + 1ll * cx[x] * 1ll * cy[y] % mod * 1ll *
                                       cx[x1] % mod * 1ll * cy[y1] % mod * 1ll *
                                       cx[x2] % mod * 1ll * cy[y2]) %
                      mod;
            }
          }
        }
      }
    }
  }
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= m; j++) {
      if (!i && !j)
        ans = (1ll * ans - 1ll * (n + 1) * 1ll * (m + 1) + mod) % mod;
      else if (!i || !j)
        ans = (1ll * ans -
               1ll * (n - i + 1) * 1ll * (m - j + 1) % mod * 1ll * (i | j) %
                   mod * 6ll +
               6ll * mod) %
              mod;
      else
        ans = (1ll * ans -
               1ll * (n - i + 1) * 1ll * (m - j + 1) % mod * 1ll * gcd(i, j) %
                   mod * 12ll +
               12ll * mod) %
              mod;
    }
  }
  printf("%d\n", ans);
  return 0;
}