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>
const long long inf = 0x3f3f3f3f3f3f3f3LL;
const long long mod = (long long)1e9 + 7;
using namespace std;
template <class T>
void smin(T& a, T val) {
  if (a > val) a = val;
}
template <class T>
void smax(T& a, T val) {
  if (a < val) a = val;
}
const long long N = 5 * (long long)1e5 + 10;
long long a, b;
long long get_sum(long long x) {
  long long tt = (x * (x - 1) / 2) % mod;
  return tt;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> a >> b;
  long long sa_1 = get_sum(a + 1);
  long long sb = get_sum(b);
  long long ans = (b * (sb % mod * sa_1 % mod) % mod) % mod;
  ans += (sb * a) % mod;
  ans %= mod;
  cout << ans;
  return 0;
}
