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;
const int inf = (1 << 30), mod = 1000000007;
const int N = 100005;
long long a, b;
long long binpow(long long x, int pw) {
  long long res = 1;
  while (pw) {
    if (pw & 1) res = res * x % mod;
    x = x * x % mod;
    pw >>= 1;
  }
  return res;
}
int main() {
  scanf("%I64d %I64d\n", &a, &b);
  long long inv2 = binpow(2ll, mod - 2);
  long long ans = a * b % mod * (b - 1) % mod * inv2 % mod *
                  (1 + b * (a + 1) % mod * inv2 % mod) % mod;
  printf("%I64d\n", ans % mod);
  return 0;
}
