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 comment(linker, "/STACK:256000000")
using namespace std;
const int P = 1000000007;
int n, m;
int get(int n, int mode) {
  if (mode == 0) {
    return 1 + (n / 2);
  } else {
    return (n + 1) / 2;
  }
}
int gcd(int x, int y) {
  if (x == 0 || y == 0) {
    return x + y;
  }
  if (x > y) {
    return gcd(x % y, y);
  } else {
    return gcd(y % x, x);
  }
}
long long cnk2(long long n) { return ((n * (n - 1)) / 2LL) % P; }
const int maxN = 4003;
int ways[maxN];
int p[maxN][maxN];
void add(int dx, int dy) {}
int main() {
  for (int i = 0; i < maxN; ++i) {
    for (int j = 0; j < maxN; ++j) {
      if (i == 0 || j == 0) {
        p[i][j] = i + j;
      } else {
        if (i > j) {
          p[i][j] = p[i - j][j];
        } else {
          p[i][j] = p[i][j - i];
        }
      }
    }
  }
  cin >> n >> m;
  long long res = 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) {
              int s = gcd(x1 + x2, y1 + y2) + gcd(x2 + x3, y2 + y3) +
                      gcd(x1 + x3, y1 + y3);
              if (s % 2 == 0) {
                long long A = get(n, x1) * get(m, y1);
                long long B = get(n, x2) * get(m, y2);
                long long C = get(n, x3) * get(m, y3);
                res += ((A * B) % P) * C;
                res %= P;
              }
            }
          }
        }
      }
    }
  }
  res -= (long long)(n + 1) * (long long)(m + 1);
  res = ((res % P) + P) % P;
  res -= 6LL * cnk2((n + 1) * (m + 1));
  res = ((res % P) + P) % P;
  for (int dx = 1; dx <= n; ++dx) {
    for (int dy = 1; dy <= m; ++dy) {
      if (dx == 0 && dy <= 0) {
        continue;
      }
      int k = p[dx][dy] - 1;
      ways[k] += (n - dx + 1) * (m - dy + 1);
      if (ways[k] >= P) {
        ways[k] -= P;
      }
    }
  }
  for (int i = 1; i < maxN; ++i) {
    res -= 12LL * (long long)(i) * (long long)(ways[i]);
    res %= P;
  }
  memset(ways, 0, sizeof(ways));
  for (int dx = 0; dx <= n; ++dx) {
    int l = 1, r = m;
    if (dx > 0) {
      l = 0;
      r = 0;
    }
    for (int dy = l; dy <= r; ++dy) {
      if (dx == 0 && dy <= 0) {
        continue;
      }
      int k = p[dx][dy] - 1;
      ways[k] += (n - dx + 1) * (m - dy + 1);
      if (ways[k] >= P) {
        ways[k] -= P;
      }
    }
  }
  for (int i = 1; i < maxN; ++i) {
    res -= 6LL * (long long)(i) * (long long)(ways[i]);
    res %= P;
  }
  res = ((res % P) + P) % P;
  cout << res << endl;
  return 0;
}