#include <bits/stdc++.h>
const double eps = 1e-9;
const double pi = acos(-1);
const int oo = 1000000000;
const int mod = 1000000007;
const double E = 2.7182818284590452353602874713527;
using namespace std;
int gcd[4002][4002];
int main() {
  int n, m, ans = 0;
  cin >> n >> m;
  n++;
  m++;
  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++)
              if (!((((x2 - x1) * (y3 - y1) - (x3 - x1) * (y2 - y1)) > 0
                         ? ((x2 - x1) * (y3 - y1) - (x3 - x1) * (y2 - y1))
                         : (-((x2 - x1) * (y3 - y1) - (x3 - x1) * (y2 - y1)))) &
                    1)) {
                int a = ((n + x1) / 2) * ((m + y1) / 2),
                    b = ((n + x2) / 2) * ((m + y2) / 2),
                    c = ((n + x3) / 2) * ((m + y3) / 2);
                if (x2 == x1 && y2 == y1) b--;
                if (x3 == x1 && y3 == y1) c--;
                if (x3 == x2 && y3 == y2) c--;
                ans = (ans + 1LL * a * b % mod * c) % mod;
              }
  for (int i = 0; i <= max(n, m); i++) {
    gcd[i][0] = i;
    for (int j = 1; j < i; j++) gcd[i][j] = gcd[j][i % j];
    gcd[i][i] = i;
  }
  for (int i = 0; i <= max(n, m); i++)
    for (int j = i + 1; j <= max(n, m); j++) gcd[i][j] = gcd[j][i];
  for (int i = 1; i < n; i++)
    for (int j = 1; j < m; j++)
      ans =
          (ans - 12LL * (gcd[i][j] - 1) * (n - i) % mod * (m - j) % mod + mod) %
          mod;
  for (int i = 1; i < n; i++)
    ans = (ans - 6LL * (gcd[i][0] - 1) * (n - i) % mod * m % mod + mod) % mod;
  for (int i = 1; i < m; i++)
    ans = (ans - 6LL * (gcd[i][0] - 1) * (m - i) % mod * n % mod + mod) % mod;
  cout << ans << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve 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>
const double eps = 1e-9;
const double pi = acos(-1);
const int oo = 1000000000;
const int mod = 1000000007;
const double E = 2.7182818284590452353602874713527;
using namespace std;
int gcd[4002][4002];
int main() {
  int n, m, ans = 0;
  cin >> n >> m;
  n++;
  m++;
  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++)
              if (!((((x2 - x1) * (y3 - y1) - (x3 - x1) * (y2 - y1)) > 0
                         ? ((x2 - x1) * (y3 - y1) - (x3 - x1) * (y2 - y1))
                         : (-((x2 - x1) * (y3 - y1) - (x3 - x1) * (y2 - y1)))) &
                    1)) {
                int a = ((n + x1) / 2) * ((m + y1) / 2),
                    b = ((n + x2) / 2) * ((m + y2) / 2),
                    c = ((n + x3) / 2) * ((m + y3) / 2);
                if (x2 == x1 && y2 == y1) b--;
                if (x3 == x1 && y3 == y1) c--;
                if (x3 == x2 && y3 == y2) c--;
                ans = (ans + 1LL * a * b % mod * c) % mod;
              }
  for (int i = 0; i <= max(n, m); i++) {
    gcd[i][0] = i;
    for (int j = 1; j < i; j++) gcd[i][j] = gcd[j][i % j];
    gcd[i][i] = i;
  }
  for (int i = 0; i <= max(n, m); i++)
    for (int j = i + 1; j <= max(n, m); j++) gcd[i][j] = gcd[j][i];
  for (int i = 1; i < n; i++)
    for (int j = 1; j < m; j++)
      ans =
          (ans - 12LL * (gcd[i][j] - 1) * (n - i) % mod * (m - j) % mod + mod) %
          mod;
  for (int i = 1; i < n; i++)
    ans = (ans - 6LL * (gcd[i][0] - 1) * (n - i) % mod * m % mod + mod) % mod;
  for (int i = 1; i < m; i++)
    ans = (ans - 6LL * (gcd[i][0] - 1) * (m - i) % mod * n % mod + mod) % mod;
  cout << ans << endl;
  return 0;
}

```