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;
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
const double eps = 1e-10;
const double pi = acos(-1.0);
const long long INF = 1LL << 62;
const int inf = 1 << 29;
const long long mod = 1000000007;
long long sum(long long n) {
  if (n & 1)
    return (n + 1) / 2 * n % mod;
  else
    return n / 2 * (n + 1) % mod;
}
int main(void) {
  long long a, b;
  cin >> a >> b;
  long long ans = 0;
  long long s = sum(a);
  for (long long i = 1; i < b; ++i) {
    long long d = (i * b) % mod * s % mod;
    long long m = (i * a) % mod;
    ans = (ans + d + m + mod) % mod;
  }
  cout << ans << endl;
  return 0;
}
