Dreamoon loves summing up something for no reason. One day he obtains two integers a and b occasionally. He wants to calculate the sum of all nice integers. Positive integer x is called nice if <image> and <image>, where k is some integer number in range [1, a].

By <image> we denote the quotient of integer division of x and y. By <image> we denote the remainder of integer division of x and y. You can read more about these operations here: http://goo.gl/AcsXhT.

The answer may be large, so please print its remainder modulo 1 000 000 007 (109 + 7). Can you compute it faster than Dreamoon?

Input

The single line of the input contains two integers a, b (1 ≤ a, b ≤ 107).

Output

Print a single integer representing the answer modulo 1 000 000 007 (109 + 7).

Examples

Input

1 1


Output

0


Input

2 2


Output

8

Note

For the first sample, there are no nice integers because <image> is always zero.

For the second sample, the set of nice integers is {3, 5}.
#include <bits/stdc++.h>
using namespace std;
long long pow(long long a, long long b, long long MOD) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y);
      if (x > MOD) x %= MOD;
    }
    y = (y * y);
    if (y > MOD) y %= MOD;
    b /= 2;
  }
  return x;
}
long long modInverse(long long a, long long m) { return pow(a, m - 2, m); }
int main(int argc, const char* argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long a, b, p, q, inv;
  long long ans;
  cin >> a >> b;
  if (b == 1) {
    cout << "0\n";
    return 0;
  } else {
    inv = modInverse(2, 1000000007LL);
    p = (((b - 1) * b) / 2) % 1000000007LL;
    q = (a + ((((b * a) % 1000000007LL) * (a + 1)) % 1000000007LL) * inv) %
        1000000007LL;
    ans = p * q;
    ans = ans % 1000000007LL;
  }
  cout << ans % 1000000007LL << "\n";
  return 0;
}
