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.math.BigInteger;
import java.util.Scanner;

public class Problem {
    public static void main(String[] args) {
        Scanner inp = new Scanner(System.in);
        int a1 = inp.nextInt(), b1 = inp.nextInt();
        BigInteger a = BigInteger.valueOf(a1);
        BigInteger b = BigInteger.valueOf(b1);
        
        BigInteger bb = BigInteger.valueOf(b1 - 1);
        BigInteger aa = BigInteger.valueOf(a1 + 1);
        BigInteger se = a.multiply(aa).multiply(b);
        BigInteger aaa = BigInteger.valueOf(a1 * 2);
        se = se.add(aaa);
        BigInteger ans = b.multiply(bb).multiply(se);
        ans = ans.divide(BigInteger.valueOf(4));
        BigInteger MOD = BigInteger.valueOf(1000000007);
        System.out.print(ans.mod(MOD));
        //BigInteger ans = b * (b - 1) * (a * b * (a + 1) + 2 * a) / 4;
        
    }
}