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>
void printImpl(const vector<T>& coll) {
  copy(coll.begin(), coll.end(), ostream_iterator<T>(cout, " "));
  cout << endl;
}
template <class T, int N>
void printImpl(T (&coll)[N]) {
  copy(coll, coll + N, ostream_iterator<T>(cout, " "));
  cout << endl;
}
template <class Key, class Value>
void printImpl(const map<Key, Value>& data) {
  typename map<Key, Value>::const_iterator it;
  for (it = data.begin(); it != data.end(); ++it) {
    cout << it->first << ":" << it->second << endl;
  }
}
template <class T>
void printImpl(const T& data) {
  cout << data << endl;
}
const long long MOD = 1000000007LL;
int main() {
  long long a, b;
  cin >> a >> b;
  long long res = 0;
  long long tt = ((b - 1) * b) / 2;
  tt %= MOD;
  res = (a * (a + 1)) / 2;
  res %= MOD;
  res *= b;
  res %= MOD;
  res *= tt;
  res %= MOD;
  res += a * tt;
  res %= MOD;
  cout << res << endl;
  return 0;
}
