#include <bits/stdc++.h>
using namespace std;
const int MD = 1000000007;
int mod_mult(long long a, long long b) { return (a * b) % MD; }
void add(int& a, int b) {
  a += b;
  if (a >= MD) a -= MD;
}
void sub(int& a, int b) {
  a -= b;
  if (a < 0) a += MD;
}
int cx[2], cy[2];
int gcd(int a, int b) {
  while (b) {
    int c = a % b;
    a = b;
    b = c;
  }
  return a;
}
int main() {
  int n, m;
  ignore = scanf("%d %d", &n, &m);
  cx[0] = n / 2 + 1;
  cx[1] = (n + 1) / 2;
  cy[0] = m / 2 + 1;
  cy[1] = (m + 1) / 2;
  int ans = 0;
  int c = 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 cp = (x2 - x1) * (y3 - y1) - (y2 - y1) * (x3 - x1);
              if (abs(cp) % 2 == 0) {
                int tmp = 1;
                tmp = mod_mult(tmp, cx[x1]);
                tmp = mod_mult(tmp, cy[y1]);
                tmp = mod_mult(tmp, cx[x2]);
                tmp = mod_mult(tmp, cy[y2]);
                tmp = mod_mult(tmp, cx[x3]);
                tmp = mod_mult(tmp, cy[y3]);
                add(ans, tmp);
                c++;
              }
            }
          }
        }
      }
    }
  }
  for (int dx = 0; dx <= n; dx++) {
    for (int dy = 0; dy <= m; dy++) {
      int a = n - dx + 1;
      int b = m - dy + 1;
      int tmp = mod_mult(a, b);
      if (dx == 0 && dy == 0) {
        sub(ans, tmp);
        continue;
      }
      if (dx > 0 && dy > 0) tmp = mod_mult(tmp, 2);
      int t = 6;
      add(t, mod_mult(6, gcd(dx, dy) - 1));
      tmp = mod_mult(tmp, t);
      sub(ans, tmp);
    }
  }
  printf("%d\n", ans);
}

### Prompt

Please provide a CPP coded solution 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;
const int MD = 1000000007;
int mod_mult(long long a, long long b) { return (a * b) % MD; }
void add(int& a, int b) {
  a += b;
  if (a >= MD) a -= MD;
}
void sub(int& a, int b) {
  a -= b;
  if (a < 0) a += MD;
}
int cx[2], cy[2];
int gcd(int a, int b) {
  while (b) {
    int c = a % b;
    a = b;
    b = c;
  }
  return a;
}
int main() {
  int n, m;
  ignore = scanf("%d %d", &n, &m);
  cx[0] = n / 2 + 1;
  cx[1] = (n + 1) / 2;
  cy[0] = m / 2 + 1;
  cy[1] = (m + 1) / 2;
  int ans = 0;
  int c = 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 cp = (x2 - x1) * (y3 - y1) - (y2 - y1) * (x3 - x1);
              if (abs(cp) % 2 == 0) {
                int tmp = 1;
                tmp = mod_mult(tmp, cx[x1]);
                tmp = mod_mult(tmp, cy[y1]);
                tmp = mod_mult(tmp, cx[x2]);
                tmp = mod_mult(tmp, cy[y2]);
                tmp = mod_mult(tmp, cx[x3]);
                tmp = mod_mult(tmp, cy[y3]);
                add(ans, tmp);
                c++;
              }
            }
          }
        }
      }
    }
  }
  for (int dx = 0; dx <= n; dx++) {
    for (int dy = 0; dy <= m; dy++) {
      int a = n - dx + 1;
      int b = m - dy + 1;
      int tmp = mod_mult(a, b);
      if (dx == 0 && dy == 0) {
        sub(ans, tmp);
        continue;
      }
      if (dx > 0 && dy > 0) tmp = mod_mult(tmp, 2);
      int t = 6;
      add(t, mod_mult(6, gcd(dx, dy) - 1));
      tmp = mod_mult(tmp, t);
      sub(ans, tmp);
    }
  }
  printf("%d\n", ans);
}

```