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}.
#d = div (x,b)
#m = mod (x,b)
#x = d*b + m = kmb + m
#значит m любое, k любое, им всегда найдётся соответствующее x.
#m может меняться от b-1, k может меняться от 1 до a
#sum (x) = sum(1..b-1) * m * sum (1..a) * k * b + sum (1..b-1)*m
#sum (x) = sum (1..a) sum (1..b-1) просуммировать: kmb + m
#sum (x) = sum (b-1) * m * sum (1..a) просуммировать: bk + 1
#S(b-1) * (b * S(a) + a)

def S (n):
    return n*(n+1)//2
#
string = input ()
string_list = string.split ()
a,b = int (string_list[0]),int (string_list[1])
itog = S(b-1)* (S(a)*b + a)
itog = int (itog % 1000000007)
print (itog)