#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000007LL;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  int w, h;
  long long s = 0;
  scanf("%d%d", &w, &h);
  ++w;
  ++h;
  for (int x0 = 0; x0 < 2; ++x0) {
    for (int y0 = 0; y0 < 2; ++y0) {
      int z0 = ((w + 1 - x0) / 2) * ((h + 1 - y0) / 2);
      for (int x1 = 0; x1 < 2; ++x1) {
        for (int y1 = 0; y1 < 2; ++y1) {
          int z1 = ((w + 1 - x1) / 2) * ((h + 1 - y1) / 2);
          for (int x2 = 0; x2 < 2; ++x2) {
            for (int y2 = 0; y2 < 2; ++y2) {
              int z2 = ((w + 1 - x2) / 2) * ((h + 1 - y2) / 2);
              if (((x1 - x0) * (y2 - y0) - (x2 - x0) * (y1 - y0)) % 2 == 0) {
                s += 1LL * z0 * z1 % MOD * z2 % MOD;
              }
            }
          }
        }
      }
    }
  }
  for (int i = 1; i < w; ++i) {
    for (int j = 1; j < h; ++j) {
      s -= 12LL * (w - i) * (h - j) * gcd(i, j) % MOD;
    }
  }
  s -= 1LL * w * w * w % MOD * h % MOD;
  s -= 1LL * h * h * h % MOD * w % MOD;
  s += 1LL * w * h % MOD;
  s = (s % MOD + MOD) % MOD;
  printf("%d\n", (int)s);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 long long MOD = 1000000007LL;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  int w, h;
  long long s = 0;
  scanf("%d%d", &w, &h);
  ++w;
  ++h;
  for (int x0 = 0; x0 < 2; ++x0) {
    for (int y0 = 0; y0 < 2; ++y0) {
      int z0 = ((w + 1 - x0) / 2) * ((h + 1 - y0) / 2);
      for (int x1 = 0; x1 < 2; ++x1) {
        for (int y1 = 0; y1 < 2; ++y1) {
          int z1 = ((w + 1 - x1) / 2) * ((h + 1 - y1) / 2);
          for (int x2 = 0; x2 < 2; ++x2) {
            for (int y2 = 0; y2 < 2; ++y2) {
              int z2 = ((w + 1 - x2) / 2) * ((h + 1 - y2) / 2);
              if (((x1 - x0) * (y2 - y0) - (x2 - x0) * (y1 - y0)) % 2 == 0) {
                s += 1LL * z0 * z1 % MOD * z2 % MOD;
              }
            }
          }
        }
      }
    }
  }
  for (int i = 1; i < w; ++i) {
    for (int j = 1; j < h; ++j) {
      s -= 12LL * (w - i) * (h - j) * gcd(i, j) % MOD;
    }
  }
  s -= 1LL * w * w * w % MOD * h % MOD;
  s -= 1LL * h * h * h % MOD * w % MOD;
  s += 1LL * w * h % MOD;
  s = (s % MOD + MOD) % MOD;
  printf("%d\n", (int)s);
  return 0;
}

```