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}.
'''

ll a,b,i,j,k,l;
	cin>>a>>b;

	ll s=0,v=b-1;
	ll x=((a*(a+1))/2)%mod;
	s=(((b*x)%mod)*(((v*(v+1))/2)%mod))%mod;

	s=(s%mod+(a*((v*(v+1))/2)%mod)%mod)%mod;
	
	cout<<s;
'''
s=raw_input().split()
a=int(s[0])
b=int(s[1])
mod=1000000007
v=b-1;
x=(a*(a+1))/2;
s=b*x*((v*(v+1))/2)

s+=a*((v*(v+1))/2)
s=s%mod;
print(s)