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.io.*;
import java.util.*;

public class P3 {
	public static void main (String[] args) {
		Scanner sc = new Scanner (System.in);
		PrintStream op = System.out;
		int a = sc.nextInt(), b = sc.nextInt();
		sc.close();
		if (b == 1) {
			op.println(0);
			return;
		}
		long sum = 0;
		int M = 1000000007;
		for (int i = 1; i < b; i++) {
			long n = a;
			long x = n % M;
			x *= i;
			x =  x % M;
			x *= b + 1;
			x =  x % M;
			sum += x;
			sum = sum % M;
			x = n;
			long y = n - 1;
			if (n % 2 == 1) {
				x = n - 1;
				y = n;
			}
			x = x >> 1;
			x *= y;
			x = x % M;
			x *= b;
			x = x % M;
			x *= i;
			x = x % M;
			sum += x;
			sum = sum % M;			
		}
		op.println(sum);
	}
}
