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}.
//package CF;

import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class A {
    static final int mod = (int)1e9+7;
    public static void main(String[] args) throws Exception
    {
	Scanner bf = new Scanner(System.in);
	PrintWriter out = new PrintWriter(System.out);
	long a = bf.nextInt(), b = bf.nextInt();
	out.println((((b*(b-1)/2)%mod)*1L*((a+(b*((a*(a+1)/2)%mod))%mod)%mod))%mod);
	out.flush();
	out.close();
    }

    static class Scanner {
	StringTokenizer st;
	BufferedReader br;

	public Scanner(InputStream s)
	{
	    br = new BufferedReader(new InputStreamReader(s));
	}

	public Scanner(FileReader fileReader)
	{
	    br = new BufferedReader(fileReader);
	}

	public String next() throws IOException
	{
	    while (st == null || !st.hasMoreTokens())
		st = new StringTokenizer(br.readLine());
	    return st.nextToken();
	}

	public int nextInt() throws IOException
	{
	    return Integer.parseInt(next());
	}

	public long nextLong() throws IOException
	{
	    return Long.parseLong(next());
	}

	public String nextLine() throws IOException
	{
	    return br.readLine();
	}

	public boolean ready() throws IOException
	{
	    return br.ready();
	}
    }
}