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 long long MOD = 1000000007;
const int MAXN = 4100;
int gcf(int a, int b) { return ((b == 0) ? a : gcf(b, a % b)); }
int W, H;
int nw[2], nh[2];
long long ans;
int main() {
  ans = 0;
  cin >> W >> H;
  nw[0] = W / 2 + 1;
  nw[1] = W + 1 - nw[0];
  nh[0] = H / 2 + 1;
  nh[1] = H + 1 - nh[0];
  for (int i = 0; i < 64; i++) {
    int par[6];
    for (int j = 0; j < 6; j++) par[j] = ((i & (1 << j)) ? 1 : 0);
    int lsum = 0;
    for (int j = 0; j < 3; j++)
      for (int k = 3; k < 6; k++) {
        if (k == j + 3) continue;
        lsum += par[j] * par[k];
      }
    if (lsum % 2 == 0) {
      long long nval = 1;
      for (int j = 0; j < 3; j++) nval = (nval * nw[par[j]]) % MOD;
      for (int j = 3; j < 6; j++) nval = (nval * nh[par[j]]) % MOD;
      ans = (ans + nval) % MOD;
    }
  }
  for (int i = 0; i <= W; i++)
    for (int j = 0; j <= H; j++) {
      if (i + j == 0) continue;
      ans = (ans + MOD -
             ((6LL * ((i * j > 0) ? 2LL : 1LL) * ((long long)W - i + 1) *
               ((long long)H - j + 1) * ((long long)gcf(i, j))) %
              MOD)) %
            MOD;
    }
  ans = (ans + MOD - (W + 1) * (H + 1)) % MOD;
  cout << ans << "\n";
  return 0;
}