#include <bits/stdc++.h>
using namespace std;
const int N = 4005;
const int MOD = 1000000007;
int w, h;
long long cnt[2][2], res, tmp, zero, t;
int gcd[N][N];
int main() {
  cin >> w >> h;
  for (int i = 0; i <= 1; i++)
    for (int j = 0; j <= 1; j++)
      cnt[i][j] = ((i == 0) ? w / 2 + 1 : w - w / 2) *
                  ((j == 0) ? h / 2 + 1 : h - h / 2);
  for (int x1 = 0; x1 <= 1; x1++)
    for (int y1 = 0; y1 <= 1; y1++)
      for (int x2 = 0; x2 <= 1; x2++)
        for (int y2 = 0; y2 <= 1; y2++)
          for (int x3 = 0; x3 <= 1; x3++)
            for (int y3 = 0; y3 <= 1; y3++) {
              int S = abs((x2 - x1) * (y3 - y1) - (x3 - x1) * (y2 - y1));
              if (S % 2 == 0) {
                tmp = (((cnt[x1][y1] * cnt[x2][y2]) % MOD) * cnt[x3][y3]) % MOD;
                res = (res + tmp) % MOD;
              }
            }
  tmp = (w + 1) * (h + 1);
  zero = (tmp + tmp * (tmp - 1) * 3) % MOD;
  ;
  for (int i = 0; i <= w; i++)
    for (int j = 0; j <= h; j++) {
      if (i == 0)
        gcd[i][j] = j;
      else if (j == 0)
        gcd[i][j] = i;
      else if (i > j)
        gcd[i][j] = gcd[i - j][j];
      else
        gcd[i][j] = gcd[i][j - i];
    }
  for (int dx = 0; dx <= w; dx++)
    for (int dy = 0; dy <= h; dy++) {
      t = gcd[dx][dy] - 1;
      if (t <= 0) continue;
      tmp = (w - dx + 1) * (h - dy + 1) * ((dx == 0 || dy == 0) ? 1 : 2);
      tmp %= MOD;
      zero = (zero + t * tmp * 6) % MOD;
    }
  res = (res - zero + MOD) % MOD;
  cout << res << endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 int N = 4005;
const int MOD = 1000000007;
int w, h;
long long cnt[2][2], res, tmp, zero, t;
int gcd[N][N];
int main() {
  cin >> w >> h;
  for (int i = 0; i <= 1; i++)
    for (int j = 0; j <= 1; j++)
      cnt[i][j] = ((i == 0) ? w / 2 + 1 : w - w / 2) *
                  ((j == 0) ? h / 2 + 1 : h - h / 2);
  for (int x1 = 0; x1 <= 1; x1++)
    for (int y1 = 0; y1 <= 1; y1++)
      for (int x2 = 0; x2 <= 1; x2++)
        for (int y2 = 0; y2 <= 1; y2++)
          for (int x3 = 0; x3 <= 1; x3++)
            for (int y3 = 0; y3 <= 1; y3++) {
              int S = abs((x2 - x1) * (y3 - y1) - (x3 - x1) * (y2 - y1));
              if (S % 2 == 0) {
                tmp = (((cnt[x1][y1] * cnt[x2][y2]) % MOD) * cnt[x3][y3]) % MOD;
                res = (res + tmp) % MOD;
              }
            }
  tmp = (w + 1) * (h + 1);
  zero = (tmp + tmp * (tmp - 1) * 3) % MOD;
  ;
  for (int i = 0; i <= w; i++)
    for (int j = 0; j <= h; j++) {
      if (i == 0)
        gcd[i][j] = j;
      else if (j == 0)
        gcd[i][j] = i;
      else if (i > j)
        gcd[i][j] = gcd[i - j][j];
      else
        gcd[i][j] = gcd[i][j - i];
    }
  for (int dx = 0; dx <= w; dx++)
    for (int dy = 0; dy <= h; dy++) {
      t = gcd[dx][dy] - 1;
      if (t <= 0) continue;
      tmp = (w - dx + 1) * (h - dy + 1) * ((dx == 0 || dy == 0) ? 1 : 2);
      tmp %= MOD;
      zero = (zero + t * tmp * 6) % MOD;
    }
  res = (res - zero + MOD) % MOD;
  cout << res << endl;
  return 0;
}

```