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.util.*;
import java.util.List;
import java.awt.*;
import java.io.*;
import java.math.*;
public class Tests{
	static Scanner in = new Scanner (System.in);
	static PrintWriter out = new PrintWriter (System.out);
	static long mod =  (long)1000000007;
  public static void main (String []args) throws NumberFormatException, IOException {
	  long a = in.nextLong(); long b = in.nextLong();
	  /* Given a,b
	   * let x = b*q + r
	   * and r != 0 and q/r = k , k belongs to [1:a] and k is uniqe for all x 
	   * => r belongs to [1:b-1] && q = r*k
	   * => x = b*(r*k) + r => x = r*(b*k + 1)
	   * => for all x = Sum from m = 1 to j = b-1 (r*(b*k + 1)) = (b*k+1)*((b*(b-1))/2)
	   * => Sum of x = Sum from i = 1 to j =a ((b*k+1)*((b*(b-1))/2))
	   */
	  long sum = 0 ;
	  for(int k = 1 ; k<=a ; k++){
		  long i = ((b*(b-1))/2);
		  i%=mod;
		  long j = (b*k+1);
		  j %= mod;
		  sum += (i*j);
		  sum %= mod;
	  }

	  out.println(sum%mod);
	  
	out.flush();
	out.close();
  }
  }
