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 void maz(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
inline void miz(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
inline int max(int a, int b) { return a > b ? a : b; }
void RI() {}
template <typename... T>
void RI(int &head, T &...tail) {
  scanf("%d", &head);
  RI(tail...);
}
const int N = 100100;
const double EPS = 1e-6;
int n, m, k;
int A[N];
vector<int> V[N];
const long long mod = 1000000007;
int main() {
  int A, B;
  RI(A, B);
  long long a = A, b = B;
  long long ans = b * (b - 1) / 2 % mod;
  ans = ans * a % mod;
  long long m = (a * (a + 1) / 2) % mod;
  long long n = (b * (b - 1) / 2) % mod;
  n = n * b % mod;
  m = n * m % mod;
  ans = (ans + m) % mod;
  printf("%d\n", (int)ans);
  return 0;
}
