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>
const int mod = 1000000007;
int a, b;
inline int getint();
inline void putint(int);
inline int mul(int, int);
inline int sum(int);
int main() {
  a = getint(), b = getint();
  if (b == 1)
    putint(0);
  else {
    int ans = sum(a);
    ans = (mul(ans, b) + a) % mod;
    ans = mul(ans, sum(b - 1));
    putint(ans);
  }
  return 0;
}
inline int getint() {
  register int num = 0;
  register char ch;
  do ch = getchar();
  while (ch < '0' || ch > '9');
  do num = num * 10 + ch - '0', ch = getchar();
  while (ch >= '0' && ch <= '9');
  return num;
}
inline void putint(int num) {
  char stack[12];
  register int top = 0;
  if (num == 0) stack[top = 1] = '0';
  for (; num; num /= 10) stack[++top] = num % 10 + '0';
  for (; top; top--) putchar(stack[top]);
  putchar('\n');
}
inline int mul(int a, int b) { return static_cast<int>(1LL * a * b % mod); }
inline int sum(int a) {
  return (a & 1) ? (mul((a + 1) >> 1, a)) : (mul(a >> 1, a + 1));
}
