Codeforces Round #272 (Div. 2) C. Dreamoon and Sums 数学

C. Dreamoon and Sums

time limit per test
1.5 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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

and

, where k is some integer number in range[1, a].

By

we denote the quotient of integer division of x and y. By

we denote the remainder of integer division of x andy. 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

is always zero.

For the second sample, the set of nice integers is {3, 5}.

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=1e5+10,M=1e6+10,inf=1e9+10,mod=1e9+7;
int main()
{
ll a,b;
scanf("%lld%lld",&a,&b);
ll ans=0;
for(ll i=1;i<b;i++)
{
ans=ans+(a*((i*b+i)%mod))%mod+(((b*i)%mod)*((a*(a- 1)/2)%mod))%mod;
ans%=mod;
}
printf("%lld\n",ans);
return 0;
}