#include <bits/stdc++.h>
using namespace std;
int G[4 << 10][4 << 10];
int gcd(int a, int b) {
  int& res = G[a][b];
  if (res != -1) return res;
  if (a == 0) return res = b;
  return res = gcd(b % a, a);
}
int main() {
  int w, h;
  scanf("%d%d", &w, &h);
  ++w;
  ++h;
  long long res = 0;
  int ax, bx, cx, ay, by, cy;
  for (ax = (0); ax < (2); ++ax)
    for (ay = (0); ay < (2); ++ay)
      for (bx = (0); bx < (2); ++bx)
        for (by = (0); by < (2); ++by)
          for (cx = (0); cx < (2); ++cx)
            for (cy = (0); cy < (2); ++cy)
              if ((ax * by + ay * bx + bx * cy + by * cx + cx * ay + cy * ax) %
                      2 ==
                  0) {
                long long r = 1;
                r *= (w - ax + 1) / 2;
                r *= (h - ay + 1) / 2;
                r %= 1000000007;
                r *= (w - bx + 1) / 2;
                r *= (h - by + 1) / 2;
                r %= 1000000007;
                r *= (w - cx + 1) / 2;
                r *= (h - cy + 1) / 2;
                r %= 1000000007;
                res += r;
                res %= 1000000007;
              }
  long long res2 = 3 * w * h;
  res2 *= w * h - 1;
  res2 += w * h;
  res2 %= 1000000007;
  memset(G, -1, sizeof(G));
  int i, j;
  int cnt = 0;
  for (i = (0); i < (w); ++i)
    for (j = (0); j < (h); ++j)
      if (i + j != 0) {
        int g = gcd(i, j);
        if (g == 1) continue;
        long long r = 6 * (g - 1);
        if (i != 0 && j != 0) r *= 2;
        r %= 1000000007;
        r *= w - i;
        r *= h - j;
        r %= 1000000007;
        res2 += r;
        res2 %= 1000000007;
      }
  printf("%d\n", int((res + 1000000007 - res2) % 1000000007));
  return 0;
};

### Prompt

Develop a solution in CPP to the problem described below:

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;
int G[4 << 10][4 << 10];
int gcd(int a, int b) {
  int& res = G[a][b];
  if (res != -1) return res;
  if (a == 0) return res = b;
  return res = gcd(b % a, a);
}
int main() {
  int w, h;
  scanf("%d%d", &w, &h);
  ++w;
  ++h;
  long long res = 0;
  int ax, bx, cx, ay, by, cy;
  for (ax = (0); ax < (2); ++ax)
    for (ay = (0); ay < (2); ++ay)
      for (bx = (0); bx < (2); ++bx)
        for (by = (0); by < (2); ++by)
          for (cx = (0); cx < (2); ++cx)
            for (cy = (0); cy < (2); ++cy)
              if ((ax * by + ay * bx + bx * cy + by * cx + cx * ay + cy * ax) %
                      2 ==
                  0) {
                long long r = 1;
                r *= (w - ax + 1) / 2;
                r *= (h - ay + 1) / 2;
                r %= 1000000007;
                r *= (w - bx + 1) / 2;
                r *= (h - by + 1) / 2;
                r %= 1000000007;
                r *= (w - cx + 1) / 2;
                r *= (h - cy + 1) / 2;
                r %= 1000000007;
                res += r;
                res %= 1000000007;
              }
  long long res2 = 3 * w * h;
  res2 *= w * h - 1;
  res2 += w * h;
  res2 %= 1000000007;
  memset(G, -1, sizeof(G));
  int i, j;
  int cnt = 0;
  for (i = (0); i < (w); ++i)
    for (j = (0); j < (h); ++j)
      if (i + j != 0) {
        int g = gcd(i, j);
        if (g == 1) continue;
        long long r = 6 * (g - 1);
        if (i != 0 && j != 0) r *= 2;
        r %= 1000000007;
        r *= w - i;
        r *= h - j;
        r %= 1000000007;
        res2 += r;
        res2 %= 1000000007;
      }
  printf("%d\n", int((res + 1000000007 - res2) % 1000000007));
  return 0;
};

```