G - Dreamoon and Sums 数学

G - Dreamoon and Sums

Time Limit:1500MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

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 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).

Sample Input

Input

1 1

Output

0

Input

2 2

Output

8

Hint

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}.

给你此公式 ，k的范围是[1,a],然后给你a和b，让你求满足此条件的x的和。

此公式可以化简为： x=k*mod(x,b)*b+mod(x,b) 即x=(k*b+1)*mod(x,b)，k可以枚举，b已知，而mod(x,b)的取值是1~b，可以通过公式直接求出。

#include<cstring>
#include<iostream>
#include<cstdio>
using namespace std;
#define mod 1000000007
int main()
{
  long long a,b;
  while(cin>>a>>b)
{


long long ans=0;
for(long long i=1;i<=a;i++)//k
{


long long c=(i*b+1)%mod;//(k*b+1);
long long d=((b-1)*b/2)%mod;//mod(x,b);
ans=(ans+c*d)%mod;
}
printf("%lld\n",ans);
}
}