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 M = 1000000007;
int memo[4001][4001];
int gcd(int a, int b) {
  if (memo[a][b]) return memo[a][b];
  return memo[a][b] = b ? gcd(b, a % b) : a;
}
int main() {
  int w, h;
  scanf("%d%d", &w, &h);
  w++;
  h++;
  long long cnt[2][2][2][2] = {};
  for (int y = 0; y < (h); y++)
    for (int x = 0; x < (w); x++) {
      for (int y1 = 0; y1 < (2); y1++)
        for (int x1 = 0; x1 < (2); x1++)
          for (int y2 = 0; y2 < (2); y2++)
            for (int x2 = 0; x2 < (2); x2++) {
              int area = (x1 - x) * (y2 - y) - (x2 - x) * (y1 - y);
              if (area % 2 == 0) cnt[y1][x1][y2][x2]++;
            }
    }
  long long even = 0;
  for (int y1 = 0; y1 < (2); y1++)
    for (int x1 = 0; x1 < (2); x1++)
      for (int y2 = 0; y2 < (2); y2++)
        for (int x2 = 0; x2 < (2); x2++) {
          even += cnt[y1][x1][y2][x2] * ((h + 1 - y1) / 2) *
                  ((w + 1 - x1) / 2) % M * ((h + 1 - y2) / 2) *
                  ((w + 1 - x2) / 2);
          even %= M;
        }
  long long bad = 0;
  for (int y = 0; y < (h); y++)
    for (int x = 0; x < (w); x++)
      if (y > 0 || x > 0) {
        long long c = (y == 0 || x == 0 ? 1 : 2);
        bad += c * 6 * (h - y) * (w - x) * (gcd(x, y) - 1);
        bad += c * 3 * (h - y) * (w - x) * 2;
        bad %= M;
      }
  bad += h * w;
  bad %= M;
  printf("%d\n", (int)((even - bad + M) % M));
  return 0;
}