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;
template <class T>
inline T sqr(const T& x) {
  return x * x;
}
template <class T>
inline void updMin(T& a, const T& b) {
  if (b < a) a = b;
}
template <class T>
inline void updMax(T& a, const T& b) {
  if (b > a) a = b;
}
long long a, b;
const long long M = 1000000007;
int main() {
  cin >> a >> b;
  long long sum = 0;
  for (int i = 1; i <= a; i++) {
    sum += i;
    if (sum >= M) sum -= M;
  }
  long long res = 0;
  for (int z = 1; z < b; z++) {
    res += a * z;
    res += ((z * b) % M) * sum;
    res %= M;
  }
  cout << res << endl;
  return 0;
}
