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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

import static java.lang.Integer.parseInt;

/**
 * TODO: describe
 *
 * @author keks
 */
public class C {
    public static void main(String[] args) throws IOException {
        final BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        final String s = reader.readLine();
        final String[] split = s.split("\\s");
        long a = parseInt(split[0]);
        long b = parseInt(split[1]);

        final long div = 1000000007L;
        long bsum = 0;
        for (int i = 0; i < b; i++) {
            bsum = (bsum + i) % div;
        }

        long asum = 0;
        for (int i = 0; i <= a; i++) {
            asum = (asum + i) % div;
        }

        long tmp = (b * asum) % div;
        tmp = (tmp + a) % div;
        long ans = (bsum * tmp) % div;
        System.out.println(ans);
    }
}
