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;
using ll = long long;
using vi = vector<int>;
const int mod = 1000000007;
int add(int a, int b) { return (a += b) < mod ? a : a - mod; }
int sub(int a, int b) { return (a -= b) >= 0 ? a : a + mod; }
int mul(int a, int b) { return 1LL * a * b % mod; }
void adds(int& a, int b) { a = add(a, b); }
void subs(int& a, int b) { a = sub(a, b); }
void muls(int& a, int b) { a = mul(a, b); }
void maxs(int& a, int b) { a = max(a, b); }
void mins(int& a, int b) { a = min(a, b); }
int pwr(int a, ll p) {
  if (p == 0) return 1;
  if (p & 1) return mul(a, pwr(a, p - 1));
  return pwr(mul(a, a), p / 2);
}
int inv(int a) { return pwr(a, mod - 2); }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int a, b;
  cin >> a >> b;
  int c = b + 1;
  int sol = 0;
  int i2 = inv(2);
  for (int i = 1; i <= a; ++i) {
    adds(sol, mul(mul(mul(add(1, mul(i, b)), b), b - 1), i2));
    adds(c, b);
  }
  cout << sol;
}
