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 <typename T, typename U>
inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
  if (x < y) x = y;
}
template <int MOD>
struct ModInt {
  static const int Mod = MOD;
  unsigned x;
  ModInt() : x(0) {}
  ModInt(signed sig) {
    int sigt = sig % MOD;
    if (sigt < 0) sigt += MOD;
    x = sigt;
  }
  ModInt(signed long long sig) {
    int sigt = sig % MOD;
    if (sigt < 0) sigt += MOD;
    x = sigt;
  }
  int get() const { return (int)x; }
  ModInt &operator+=(ModInt that) {
    if ((x += that.x) >= MOD) x -= MOD;
    return *this;
  }
  ModInt &operator-=(ModInt that) {
    if ((x += MOD - that.x) >= MOD) x -= MOD;
    return *this;
  }
  ModInt &operator*=(ModInt that) {
    x = (unsigned long long)x * that.x % MOD;
    return *this;
  }
  ModInt &operator/=(ModInt that) { return *this *= that.inverse(); }
  ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
  ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
  ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
  ModInt operator/(ModInt that) const { return ModInt(*this) /= that; }
  ModInt inverse() const {
    long long a = x, b = MOD, u = 1, v = 0;
    while (b) {
      long long t = a / b;
      a -= t * b;
      std::swap(a, b);
      u -= t * v;
      std::swap(u, v);
    }
    return ModInt(u);
  }
};
int main() {
  int a, b;
  scanf("%d%d", &a, &b);
  ModInt<1000000007> ans = 0;
  ModInt<1000000007> t = ModInt<1000000007>(b) * (b - 1) / 2;
  for (int(k) = (int)(1); (k) <= (int)(a); ++(k)) {
    ans += (ModInt<1000000007>(k) * b + 1) * t;
  }
  printf("%d\n", ans.get());
  return 0;
}
