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.io.*;
import java.math.*;
import java.util.regex.*;

public class Task476C {
  private static long MOD = 1000000007L;
  public static void main(String[] args) throws IOException {
    BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    StringTokenizer st = new StringTokenizer(f.readLine());
    
    BigInteger a = new BigInteger(st.nextToken());
    BigInteger b = new BigInteger(st.nextToken());
    BigInteger m = new BigInteger("1000000007");
    
    BigInteger c = b.multiply(b.subtract(BigInteger.ONE))
                    .divide(new BigInteger("2")).mod(m);
    c = c.multiply(a.add(b.multiply(a.multiply
                  (a.add(BigInteger.ONE)).divide(new BigInteger("2"))))).mod(m);
    out.println(c.toString());
    
    out.close();
    System.exit(0);
  }
}