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.*;
import java.math.*;

public class Main {
    public static void main(String[] args)
    {
        Scanner cin=new Scanner(System.in);
        long a; long b;
        long MOD = 1000000007;
        a = cin.nextLong(); b = cin.nextLong(); 
        
        BigInteger t1 = BigInteger.valueOf(a + 1).multiply(BigInteger.valueOf(a)).multiply(BigInteger.valueOf(b))
                .divide(BigInteger.valueOf(2));
        t1 = t1.add(BigInteger.valueOf(a));
        BigInteger t2 = BigInteger.valueOf(b).multiply(BigInteger.valueOf(b - 1)).divide(BigInteger.valueOf(2));
        t1 = t1.mod(BigInteger.valueOf(MOD));
        t2 = t2.mod(BigInteger.valueOf(MOD));
        BigInteger ans = t1.multiply(t2);
        ans = ans.mod(BigInteger.valueOf(MOD));
        System.out.println(ans);
        cin.close();
    }       
}