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}.

import java.util.Scanner;
//Not Solved !!
public class DreamoonAndSums {
	public static void main(String[] args) {
		Scanner sc= new Scanner(System.in);
		long a = sc.nextInt();
		long b = sc.nextInt();
		
		
		long ans = 0;
		
		long mod = (long)(1e9)+7;
		long inv = inv(2,mod);
		long x = ((1l*b%mod*((b-1)%mod))%mod)*(inv%mod)%mod;
		long y = (a + (b%mod)  * (a%mod)%mod * ((a+1)%mod)%mod * inv%mod);
		
		ans = 1l*x%mod *y%mod;
		
		System.out.println(ans%mod);
		sc.close();
	}
	
	static  long inv(long x , long mod){
	    long r, y;
	    for(r = 1 , y = mod - 2 ; y != 0 ;x = x * x % mod, y>>=1)
	    	if ((y & 1) == 1)
	    		r = r * x % mod;
	    return r;
	}

}
