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}.
mod=10**9+7
a,b=map(int,raw_input().split())
if a%2!=0 and b%2!=0:
    ans=(a%mod)*(b%mod)*(((b-1)/2)%mod)
    ans+=(a%mod)*((b%mod*b%mod)%mod)*(((a+1)/2)%mod)*(((b-1)/2)%mod)
elif a%2!=0:
    ans=(a%mod)*((b/2)%mod)*((b-1)%mod)
    ans+=(a%mod)*(((b/2)%mod*(b/2)%mod)%mod)*((a+1)%mod)*((b-1)%mod)
elif b%2!=0:
    ans=(a%mod)*(b%mod)*(((b-1)/2)%mod)
    ans+=((a/2)%mod)*((b%mod*b%mod)%mod)*((a+1)%mod)*(((b-1)/2)%mod)
else:
    ans=(a%mod)*((b/2)%mod)*((b-1)%mod)
    ans+=(a%mod)*(((b/2)%mod*(b/2)%mod)%mod)*((a+1)%mod)*((b-1)%mod)
    
    

print ans%mod