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>
template <class T>
inline bool rd(T &ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) return 0;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  ret *= sgn;
  return 1;
}
template <class T>
inline void pt(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) pt(x / 10);
  putchar(x % 10 + '0');
}
using namespace std;
const long long mod = 1000000007;
long long a, b;
int main() {
  while (cin >> a >> b) {
    long long ans = (a * (1 + a)) / 2;
    ans %= mod;
    ans *= b;
    ans %= mod;
    ans += a;
    ans %= mod;
    long long B = b * (b - 1) / 2 % mod;
    ans *= B;
    ans %= mod;
    cout << ans << endl;
  }
  return 0;
}
