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;
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
void solve();
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed << setprecision(20);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
const long long M = 1e9 + 7;
long long f(long long n) { return ((n * (n + 1)) / 2) % M; }
void solve() {
  long long a, b;
  cin >> a >> b;
  long long ans = a * f(b - 1);
  ans %= M;
  long long lol = b * f(b - 1);
  lol %= M;
  for (long long i = 1; i <= a; ++i) {
    ans += (lol * i) % M;
    if (ans >= M) ans -= M;
  }
  cout << ans;
}
