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;
constexpr int INF = 1e9 + 1;
constexpr long long LLINF = 1e18 + 1;
template <typename T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
const int mod = 1000000007;
int add(int a, int b) {
  a += b;
  if (a >= mod) a -= mod;
  return a;
}
int mul(int a, int b) { return a * 1LL * b % mod; }
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long a, b;
  while (cin >> a >> b) {
    long long res = (b * (b - 1)) / 2, rest = (a * (a + 1)) / 2;
    rest %= mod;
    res %= mod;
    res = mul(res, add(mul(rest, b), a));
    printf("%lld\n", res);
  }
}
