#include <bits/stdc++.h>
using namespace std;
long long x[2];
long long y[2];
int gcd(int x, int y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
long long calc(long long x) {
  if (x == 1) return 1;
  if (x == 2) return 6;
  return 6 + 6 * (x - 2);
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  x[0] = (n + 2) / 2;
  x[1] = (n + 1) / 2;
  y[0] = (m + 2) / 2;
  y[1] = (m + 1) / 2;
  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 x3 = 0; x3 < 2; x3++)
            for (int y3 = 0; y3 < 2; y3++) {
              int now =
                  x1 * y2 + x2 * y3 + x3 * y1 + y1 * x2 + y2 * x3 + y3 * x1;
              if (now % 2 == 0) {
                long long tmp = 1;
                tmp = (tmp * x[x1]) % 1000000007;
                tmp = (tmp * y[y1]) % 1000000007;
                tmp = (tmp * x[x2]) % 1000000007;
                tmp = (tmp * y[y2]) % 1000000007;
                tmp = (tmp * x[x3]) % 1000000007;
                tmp = (tmp * y[y3]) % 1000000007;
                ans = (ans + tmp) % 1000000007;
              }
            }
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= m; j++) {
      long long base = ((n - i + 1) * (m - j + 1)) % 1000000007;
      int g;
      if (i == 0)
        g = j;
      else if (j == 0)
        g = i;
      else
        g = gcd(i, j);
      g++;
      long long tmp = (calc(g) * base) % 1000000007;
      if (i != 0 && j != 0) tmp = (tmp * 2) % 1000000007;
      ans = (ans - tmp + 1000000007) % 1000000007;
    }
  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;
long long x[2];
long long y[2];
int gcd(int x, int y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
long long calc(long long x) {
  if (x == 1) return 1;
  if (x == 2) return 6;
  return 6 + 6 * (x - 2);
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  x[0] = (n + 2) / 2;
  x[1] = (n + 1) / 2;
  y[0] = (m + 2) / 2;
  y[1] = (m + 1) / 2;
  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 x3 = 0; x3 < 2; x3++)
            for (int y3 = 0; y3 < 2; y3++) {
              int now =
                  x1 * y2 + x2 * y3 + x3 * y1 + y1 * x2 + y2 * x3 + y3 * x1;
              if (now % 2 == 0) {
                long long tmp = 1;
                tmp = (tmp * x[x1]) % 1000000007;
                tmp = (tmp * y[y1]) % 1000000007;
                tmp = (tmp * x[x2]) % 1000000007;
                tmp = (tmp * y[y2]) % 1000000007;
                tmp = (tmp * x[x3]) % 1000000007;
                tmp = (tmp * y[y3]) % 1000000007;
                ans = (ans + tmp) % 1000000007;
              }
            }
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= m; j++) {
      long long base = ((n - i + 1) * (m - j + 1)) % 1000000007;
      int g;
      if (i == 0)
        g = j;
      else if (j == 0)
        g = i;
      else
        g = gcd(i, j);
      g++;
      long long tmp = (calc(g) * base) % 1000000007;
      if (i != 0 && j != 0) tmp = (tmp * 2) % 1000000007;
      ans = (ans - tmp + 1000000007) % 1000000007;
    }
  cout << ans << endl;
  return 0;
}

```