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>
#pragma gcc optimize("O3, unroll-loops")
using namespace std;
const int mod = 1000000007;
int gcds[4001][4001];
int64_t fpow(int i, int j) {
  int64_t ans = 1;
  while (j--) ans *= i;
  return ans;
}
signed main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  int64_t n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) gcds[i][0] = i;
  for (int i = 1; i <= m; i++) gcds[0][i] = i;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++)
      gcds[i][j] = i > j ? gcds[i % j][j] : gcds[i][j % i];
  int64_t cnts = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      int g = gcds[i][j];
      int64_t targ = min((n - i) / (i / g), (m - j) / (j / g));
      (cnts += targ * (targ + 1) * (2 * targ + 1) / 6 * (i / g) * (j / g) +
               (n - i + 1) * (m - j + 1) * targ -
               targ * (targ + 1) / 2 *
                   ((n - i + 1) * (j / g) + (m - j + 1) * (i / g))) %= mod;
    }
  cnts = (cnts * 2 * 6 + (n + 1) * (m + 1) * (m * (m - 1) + n * (n - 1)) +
          (n + 1) * (m + 1) * ((n + 1) * (m + 1) - 1) * 3 + (n + 1) * (m + 1)) %
         mod;
  int64_t ans = 0;
  for (int i = 0; i < (1 << 6); i++)
    if (~(i >> 2 & i >> 4 ^ i >> 0 & i >> 5 ^ i >> 1 & i >> 3 ^
          i >> 2 & i >> 3 ^ i >> 0 & i >> 4 ^ i >> 1 & i >> 5) &
        1)
      ans += fpow((n + 1) / 2, __builtin_popcount(i & 7)) % mod *
             fpow((n + 2) / 2, 3 - __builtin_popcount(i & 7)) % mod *
             fpow((m + 1) / 2, __builtin_popcount(i >> 3)) % mod *
             fpow((m + 2) / 2, 3 - __builtin_popcount(i >> 3)) % mod;
  cout << ((ans - cnts) % mod + mod) % mod << endl;
}