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 mod = 1e9 + 7;
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long modInverse(long long n, long long p) { return power(n, p - 2, p); }
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long a, b;
  cin >> a >> b;
  long long ans = 0;
  for (long long i = 1; i < b; i++) {
    long long first = ((b * i) % mod + i) % mod;
    long long last = ((((b * a) % mod) * i) % mod + i) % mod;
    long long tm = a;
    long long val = ((long long)tm * (first + last) % mod) % mod;
    val = val % mod;
    ans = (ans + (long long)val) % mod;
  }
  ans = (ans * modInverse(2, mod)) % mod;
  cout << ans << endl;
  return 0;
}
