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>
inline T modinv(T a, T n) {
  T i = n, v = 0, d = 1;
  while (a > 0) {
    T t = i / a, x = a;
    a = i % x;
    i = x;
    x = d;
    d = v - t * x;
    v = x;
  }
  return (v + n) % n;
}
long long modpow(long long n, long long k, long long mod) {
  long long ans = 1;
  while (k > 0) {
    if (k & 1) ans = (ans * n) % mod;
    k >>= 1;
    n = (n * n) % mod;
  }
  return ans % mod;
}
template <class T>
string str(T Number) {
  string Result;
  ostringstream convert;
  convert << Number;
  Result = convert.str();
  return Result;
}
int StringToNumber(const string &Text) {
  istringstream ss(Text);
  int result;
  return ss >> result ? result : 0;
}
template <class T>
inline vector<pair<T, int> > FACTORISE(T n) {
  vector<pair<T, int> > R;
  for (T i = 2; n > 1;) {
    if (n % i == 0) {
      int C = 0;
      for (; n % i == 0; C++, n /= i)
        ;
      R.push_back(make_pair(i, C));
    }
    i++;
    if (i > n / i) i = n;
  }
  if (n > 1) R.push_back(make_pair(n, 1));
  return R;
}
template <class T>
inline T TOTIENT(T n) {
  vector<pair<T, int> > R = FACTORISE(n);
  T r = n;
  for (int i = 0; i < R.size(); i++) r = r / R[i].first * (R[i].first - 1);
  return r;
}
template <class T>
inline T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
double rnd(float d) { return floor(d + 0.49); }
int main() {
  long long a, b;
  cin >> a >> b;
  long long sum = 0;
  for (long long i = 1; i <= a; i++) {
    long long d =
        (((b * (b - 1) / 2) % 1000000007) * ((b * i + 1) % 1000000007)) %
        1000000007;
    sum = (sum % 1000000007 + d % 1000000007) % 1000000007;
  }
  cout << sum << endl;
  return 0;
}
