#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int n, m;
long long cx[2], cy[2], num[4];
int cal(int x, int y, int x1, int y1, int x2, int y2) {
  return ((x1 - x) * (y2 - y) - (x2 - x) * (y1 - y)) % 2;
}
void upd(long long& x, long long y) {
  x += y;
  x %= mod;
  if (x < 0) x += mod;
}
int gcd(int a, int b) { return (!b) ? a : gcd(b, a % b); }
int main() {
  scanf("%d%d", &n, &m);
  cx[0] = (n >> 1) + 1;
  cx[1] = (n + 1) >> 1;
  cy[0] = (m >> 1) + 1;
  cy[1] = (m + 1) >> 1;
  for (int i = 0; i <= 1; i++)
    for (int j = 0; j <= 1; j++) {
      num[(i << 1) + j] = cx[i] * cy[j] % mod;
    }
  long long ans = 0;
  for (int x = 0; x <= 1; x++)
    for (int y = 0; y <= 1; y++)
      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++)
              if (!cal(x, y, x1, y1, x2, y2)) {
                int p1 = (x1 << 1) + y1, p2 = (x2 << 1) + y2, p = (x << 1) + y;
                int used[4] = {};
                long long tmp = 1;
                used[p1]++, used[p2]++, used[p]++;
                for (int i = 0; i < 4; i++) {
                  if (used[i] > num[i]) {
                    tmp = 0;
                    break;
                  }
                  for (int k = 0; k < used[i]; k++) {
                    tmp = tmp * (num[i] - k) % mod;
                  }
                }
                upd(ans, tmp);
              }
  for (int i = 2; i <= n; i++)
    upd(ans, -6ll * ((n - i + 1) * (m + 1) % mod) * (i - 1) % mod);
  for (int j = 2; j <= m; j++)
    upd(ans, -6ll * ((n + 1) * (m - j + 1) % mod) * (j - 1) % mod);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      upd(ans,
          -12ll * ((n - i + 1) * (m - j + 1) % mod) * (gcd(i, j) - 1) % mod);
    }
  cout << ans << endl;
  return 0;
}

### 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 int mod = 1e9 + 7;
int n, m;
long long cx[2], cy[2], num[4];
int cal(int x, int y, int x1, int y1, int x2, int y2) {
  return ((x1 - x) * (y2 - y) - (x2 - x) * (y1 - y)) % 2;
}
void upd(long long& x, long long y) {
  x += y;
  x %= mod;
  if (x < 0) x += mod;
}
int gcd(int a, int b) { return (!b) ? a : gcd(b, a % b); }
int main() {
  scanf("%d%d", &n, &m);
  cx[0] = (n >> 1) + 1;
  cx[1] = (n + 1) >> 1;
  cy[0] = (m >> 1) + 1;
  cy[1] = (m + 1) >> 1;
  for (int i = 0; i <= 1; i++)
    for (int j = 0; j <= 1; j++) {
      num[(i << 1) + j] = cx[i] * cy[j] % mod;
    }
  long long ans = 0;
  for (int x = 0; x <= 1; x++)
    for (int y = 0; y <= 1; y++)
      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++)
              if (!cal(x, y, x1, y1, x2, y2)) {
                int p1 = (x1 << 1) + y1, p2 = (x2 << 1) + y2, p = (x << 1) + y;
                int used[4] = {};
                long long tmp = 1;
                used[p1]++, used[p2]++, used[p]++;
                for (int i = 0; i < 4; i++) {
                  if (used[i] > num[i]) {
                    tmp = 0;
                    break;
                  }
                  for (int k = 0; k < used[i]; k++) {
                    tmp = tmp * (num[i] - k) % mod;
                  }
                }
                upd(ans, tmp);
              }
  for (int i = 2; i <= n; i++)
    upd(ans, -6ll * ((n - i + 1) * (m + 1) % mod) * (i - 1) % mod);
  for (int j = 2; j <= m; j++)
    upd(ans, -6ll * ((n + 1) * (m - j + 1) % mod) * (j - 1) % mod);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      upd(ans,
          -12ll * ((n - i + 1) * (m - j + 1) % mod) * (gcd(i, j) - 1) % mod);
    }
  cout << ans << endl;
  return 0;
}

```