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.*;

public class Main {
  // main
  public static void main(String [] args) throws IOException {
    // makes the reader and writer
    BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
    
    // read in 
    StringTokenizer st = new StringTokenizer(f.readLine());
    int a = Integer.parseInt(st.nextToken());
    int b = Integer.parseInt(st.nextToken());
    
    // add
    long mod = 1000000007;
    long sums = (((((long) (b-1))*b/2)%mod)*(b+1))%mod;
    long terms = (((((long) b)*(b-1)/2)%mod)*b)%mod;
    long ks = ((((long) a)*(a-1))/2)%mod;
    long sum = (((sums*a)%mod)+(ks*terms))%mod;

    // write to out
    out.println(sum);
    
    // cleanup
    out.close();
    System.exit(0);
  }
}