#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
int gcd[4005][4005];
long long way[2][2];
int main() {
  int r, c;
  scanf("%d%d", &r, &c);
  r++;
  c++;
  int i, j, k = max(r, c);
  gcd[0][0] = 1;
  for (i = 1; i <= k; i++) {
    gcd[i][0] = gcd[0][i] = i;
    for (j = 1; j <= i; j++) gcd[i][j] = gcd[j][i] = gcd[i - j][j];
  }
  for (i = 0; i < r; i++)
    for (j = 0; j < c; j++) ++way[i & 1][j & 1];
  long long ans = 0;
  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++)
          for (int x = 0; x < 2; x++)
            for (int y = 0; y < 2; y++) {
              k = (x1 - x) * (y2 - y) - (x2 - x) * (y1 - y);
              if (k % 2 == 0)
                ans += way[x1][y1] * way[x2][y2] % mod * way[x][y] % mod;
            }
  long long s = (r * c) % mod, zero = s + 3 * s * (s - 1) % mod;
  for (int dx = 0; dx < r; dx++)
    for (int dy = 0; dy < c; dy++)
      zero += 6ll * (r - dx) * (c - dy) % mod * (gcd[dx][dy] - 1) *
              ((dx > 0 && dy > 0) ? 2 : 1) % mod;
  ans = (ans - zero % mod + mod) % mod;
  cout << ans << endl;
}

### Prompt

Generate a cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
int gcd[4005][4005];
long long way[2][2];
int main() {
  int r, c;
  scanf("%d%d", &r, &c);
  r++;
  c++;
  int i, j, k = max(r, c);
  gcd[0][0] = 1;
  for (i = 1; i <= k; i++) {
    gcd[i][0] = gcd[0][i] = i;
    for (j = 1; j <= i; j++) gcd[i][j] = gcd[j][i] = gcd[i - j][j];
  }
  for (i = 0; i < r; i++)
    for (j = 0; j < c; j++) ++way[i & 1][j & 1];
  long long ans = 0;
  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++)
          for (int x = 0; x < 2; x++)
            for (int y = 0; y < 2; y++) {
              k = (x1 - x) * (y2 - y) - (x2 - x) * (y1 - y);
              if (k % 2 == 0)
                ans += way[x1][y1] * way[x2][y2] % mod * way[x][y] % mod;
            }
  long long s = (r * c) % mod, zero = s + 3 * s * (s - 1) % mod;
  for (int dx = 0; dx < r; dx++)
    for (int dy = 0; dy < c; dy++)
      zero += 6ll * (r - dx) * (c - dy) % mod * (gcd[dx][dy] - 1) *
              ((dx > 0 && dy > 0) ? 2 : 1) % mod;
  ans = (ans - zero % mod + mod) % mod;
  cout << ans << endl;
}

```