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>
#pragma comment(linker, "/STACK:1677721600")
using namespace std;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double ee = exp(1.0);
const int inf = 0x3f3f3f3f;
const int maxn = 1e6 + 10;
const double pi = acos(-1.0);
const long long iinf = 0x3f3f3f3f3f3f3f3f;
int readT() {
  char c;
  int ret = 0, flg = 0;
  while (c = getchar(), (c < '0' || c > '9') && c != '-')
    ;
  if (c == '-')
    flg = 1;
  else
    ret = c ^ 48;
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c ^ 48);
  return flg ? -ret : ret;
}
long long readTL() {
  char c;
  int flg = 0;
  long long ret = 0;
  while (c = getchar(), (c < '0' || c > '9') && c != '-')
    ;
  if (c == '-')
    flg = 1;
  else
    ret = c ^ 48;
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c ^ 48);
  return flg ? -ret : ret;
}
int main() {
  long long a = readTL();
  long long b = readTL();
  long long ans = 0;
  for (int i = (1); i <= (a); i++) {
    ans = (ans + (i * b + 1) % mod) % mod;
  }
  long long t = 0;
  for (int i = (1); i <= (b - 1); i++) {
    t = (t + (i * ans) % mod) % mod;
  }
  cout << t << endl;
  return 0;
}
