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.math.*;
import java.util.*;


public class Main {


	public static class pair implements Comparable<pair>
	{
		int a;
		int b;
		public pair(int pa, int pb)
		{
			a = pa; b= pb;
		}
		@Override
		public int compareTo(pair o) {
			if(this.a < o.a)
				return -1;
			if(this.a > o.a)
				return 1;
			return Integer.compare(o.b, this.b);
		}
	}


	//int n = Integer.parseInt(in.readLine());
	//int n = Integer.parseInt(spl[0]);
	//String[] spl = in.readLine().split(" ");
		
	public static void main (String[] args) throws Exception
	{	
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		String[] spl = in.readLine().split(" ");
		long a = Integer.parseInt(spl[0]);
		long b  = Integer.parseInt(spl[1]);
		long mod = 1000000007;
		long pp = (b*(b-1))/2;
		long ans = 0;
		pp%=mod;
		for (long k = 1; k <= a; k++) {
			long cur = k*b+1;
			cur%=mod;
			ans+=pp*cur;
			ans%=mod;
		}
		System.out.println(ans);
	}

}