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;
long long ans;
long long n, m;
long long c[2][2];
long long gcd(long long a, long long b) {
  if (a == 0 || b == 0) return a + b;
  while ((a %= b) && (b %= a))
    ;
  return a + b;
}
int main() {
  scanf("%d%d", &n, &m);
  c[0][0] = (n / 2 + 1) * (m / 2 + 1);
  c[0][1] = (n / 2 + 1) * ((m + 1) / 2);
  c[1][0] = ((n + 1) / 2) * (m / 2 + 1);
  c[1][1] = ((n + 1) / 2) * ((m + 1) / 2);
  long long area;
  for (long long x1 = 0; x1 < 2; ++x1)
    for (long long y1 = 0; y1 < 2; ++y1)
      for (long long x2 = 0; x2 < 2; ++x2)
        for (long long y2 = 0; y2 < 2; ++y2)
          for (long long x3 = 0; x3 < 2; ++x3)
            for (long long y3 = 0; y3 < 2; ++y3) {
              area = (x2 - x1) * (y3 - y1) - (x3 - x1) * (y2 - y1);
              if (area % 2 == 0) {
                ans = ans +
                      (long long)c[x1][y1] * c[x2][y2] % mod * c[x3][y3] % mod;
                if (ans >= mod) ans -= mod;
              }
            }
  long long tmp = (n + 1) * (m + 1);
  ans = ans - (long long)tmp % mod;
  if (ans < 0) ans += mod;
  ans = ans - (long long)3 * tmp * (tmp - 1) % mod;
  if (ans < 0) ans += mod;
  for (long long i = 0; i <= n; ++i)
    for (long long j = 0; j <= m; ++j) {
      if (i == 0 && j == 0) continue;
      tmp = gcd(i, j);
      ans = ans - (long long)6 * (n - i + 1) * (m - j + 1) * (tmp - 1);
      if (ans < 0) ans += mod;
      if (i == 0 || j == 0) continue;
      ans = ans - (long long)6 * (n - i + 1) * (m - j + 1) * (tmp - 1);
      if (ans < 0) ans += mod;
    }
  printf("%d\n", ans);
  return 0;
}