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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class C {
    private static BufferedReader in;
    private static StringTokenizer st;
    private static PrintWriter out;

    public static void main(String[] args) throws NumberFormatException,
            IOException {
        in = new BufferedReader(new InputStreamReader(System.in));
        st = new StringTokenizer("");
        out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(
                System.out)));
        int a = nextInt();
        int b = nextInt();
        long summa = 0;
        long r = (long) (1e9 + 7);
        long s = (sum(a, r) * b) % r;
        for (int i = 1; i <= b - 1; i++) {
            summa += ((s + a) * i) % r;
        }
        out.println(summa % r);
        out.close();
    }

    private static long sum(long n, long mod) {
        long res = (1 + n) * n / 2;
        return res % mod;
    }

    private static long nextLong() throws NumberFormatException, IOException {
        return Long.parseLong(next());
    }

    private static int nextInt() throws NumberFormatException, IOException {
        return Integer.parseInt(next());
    }

    private static String next() throws IOException {
        while (!st.hasMoreTokens())
            st = new StringTokenizer(in.readLine());
        return st.nextToken();
    }
}
