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}.
/**
 * Created by Dima on 12.10.2014.
 */

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class TaskC272 {
    private static BufferedReader reader;
    private static PrintWriter writer;
    private static StringTokenizer st;

    public static void main(String[] args) throws IOException {
        reader = new BufferedReader(new InputStreamReader(System.in));
        writer = new PrintWriter(System.out, true);
        solve();
    }

    private static void solve() throws IOException {
        final long N = 1000000007;
        long a = nextInt();
        long b = nextInt();
        long sum_b= 0;
        long sum_a = 0;
        for(long l = 1; l <= b - 1; ++l) sum_b += l;
        sum_b = sum_b % N;
        for(long l = 1; l <= a; ++l) sum_a += l;
        long answer = sum_b * ((((b % N) * (sum_a % N)) % N + a % N) % N);
        answer = answer % N;
        writer.println(answer);
        writer.close();
    }

    private static int nextInt() throws IOException {
        return Integer.parseInt(readToken());
    }

    private static String readToken() throws IOException {
        while (st == null || !st.hasMoreElements()) st = new StringTokenizer(reader.readLine(), " ");
        return st.nextToken();
    }
}
