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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int maxn = 4444444;
long long cnt[4], c[maxn][4], w, h;
inline long long gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
long long work() {
  long long ans = 0;
  int v = 0;
  for (int i = 1; i < h + 1; i++) {
    for (int j = 1; j < w + 1; j++) {
      if ((v = gcd(i, j)) != 1) {
        ans += ((w - j + 1) * (h - i + 1) * (v - 1)) % mod;
        ans %= mod;
      }
    }
  }
  ans = ans * 2 + c[w + 1][3] * (h + 1) % mod + c[h + 1][3] * (w + 1) % mod;
  ans %= mod;
  return ans;
}
int main() {
  scanf("%d%d", &w, &h);
  for (int i = 0; i < w + 1; i++)
    for (int j = 0; j < h + 1; j++) {
      int c = i % 2 * 2 + j % 2;
      cnt[c]++;
    }
  c[0][0] = 1;
  for (int i = 1; i < maxn; i++) {
    c[i][0] = 1;
    for (int j = 1; j < 4; j++) {
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
    }
  }
  long long s = (w + 1) * (h + 1), ans = 0;
  for (int i = 0; i < 4; i++) {
    ans += c[cnt[i]][3];
    ans += c[cnt[i]][2] * (s - cnt[i]) % mod;
  }
  ans %= mod;
  ans -= work() - mod;
  ans = (ans * 6) % mod;
  printf("%lld", ans);
  return 0;
}