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.BufferedReader;
import java.io.InputStreamReader;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.util.*;

public class Main
{
    public static void main(String args[])throws Exception
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st;
        st=new StringTokenizer(br.readLine());
        long a=Long.parseLong(st.nextToken());
        long b=Long.parseLong(st.nextToken());
        BigInteger fir=BigInteger.valueOf(a).multiply(BigInteger.valueOf(a+1)).divide(BigInteger.valueOf(2));
        fir=fir.multiply(BigInteger.valueOf(b));
        fir=fir.add(BigInteger.valueOf(a)).mod(BigInteger.valueOf(1000000007));
        BigInteger sec=BigInteger.valueOf(b).multiply(BigInteger.valueOf(b-1)).divide(BigInteger.valueOf(2)).mod(BigInteger.valueOf(1000000007));
        fir=fir.multiply(sec).mod(BigInteger.valueOf(1000000007));
        System.out.print(fir);
    }
}