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 modulo = 1e9 + 7;
long long dp[4096][4096];
int G[4096][4096];
int N, M;
int cc(int x, int parity) {
  ++x;
  parity ^= 1;
  if (parity == 0) return x / 2;
  return x / 2 + x % 2;
}
void init() {
  for (int i = 0; i <= 4001; ++i) G[i][0] = i;
  for (int j = 0; j <= 4001; ++j) G[0][j] = j;
  for (int i = 1; i <= 4001; ++i)
    for (int j = 1; j <= 4001; ++j)
      if (i > j)
        G[i][j] = G[i - j][i];
      else
        G[i][j] = G[i][j - i];
}
inline int ABS(int t1) { return (t1 < 0) ? (-t1) : t1; }
int main() {
  cin >> N >> M;
  init();
  for (int i = 0; i <= N; ++i) dp[i][0] = i * (i - 1) / 2;
  for (int i = 0; i <= M; ++i) dp[0][i] = i * (i - 1) / 2;
  for (int i = 1; i <= N; ++i)
    for (int j = 1; j <= M; ++j) {
      dp[i][j] =
          dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1] + 2 * (G[i][j] - 1);
      dp[i][j] %= modulo;
    }
  for (int i = 1; i <= N; ++i) {
    dp[i][0] += dp[i - 1][0];
    dp[i][0] %= modulo;
  }
  for (int j = 1; j <= M; ++j) {
    dp[0][j] += dp[0][j - 1];
    dp[0][j] %= modulo;
  }
  for (int i = 1; i <= N; ++i)
    for (int j = 1; j <= M; ++j) {
      dp[i][j] += dp[i][j - 1] + dp[i - 1][j] - dp[i - 1][j - 1];
      dp[i][j] %= modulo;
    }
  long long all = (N + 1) * (M + 1);
  long long col = dp[N][M] * 6 + all + all * (all - 1) * 3;
  col %= modulo;
  long long triag = 0;
  int br = 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) {
              long long cx1, cx2, cx3, cy1, cy2, cy3;
              if (ABS((x1 - x2) * (y1 - y3) - (x1 - x3) * (y1 - y2)) % 2 == 1)
                continue;
              cx1 = cc(N, x1);
              cy1 = cc(M, y1);
              cx2 = cc(N, x2);
              cy2 = cc(M, y2);
              cx3 = cc(N, x3);
              cy3 = cc(M, y3);
              ++br;
              triag = (triag + ((cx1 * cx2 * cx3) % modulo) *
                                   ((cy1 * cy2 * cy3) % modulo)) %
                      modulo;
            }
  cout << (triag - col + modulo) % modulo << endl;
}