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.File;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import static java.lang.Math.*;
import static java.util.Arrays.*;

public class Main {

	public static void main(String[] args) throws IOException {
		new Main().run();
	}

	BufferedReader in;
	PrintWriter out;
	StringTokenizer st = new StringTokenizer("");

	private void run() throws IOException {
		if (new File("input.txt").exists())
			in = new BufferedReader(new FileReader("input.txt"));
		else
			in = new BufferedReader(new InputStreamReader(System.in));
		if (new File("output.txt").exists())
			out = new PrintWriter("output.txt");
		else
			out = new PrintWriter(System.out);

		solve();

		in.close();
		out.close();
	}

	final long mod = 1000 * 1000 * 1000 + 7;

	long a, b;
	long ans = 0;

	long gcd(long a, long b) {
		while (a > 0 && b > 0)
			if (a > b)
				a %= b;
			else
				b %= a;
		return a + b;
	}

	void solve() throws IOException {
		a = nextLong();
		b = nextLong();

		for (long q = 1; q < b; q++) {
//			long g = gcd(q, b);
			long f = a;
			
//			System.err.println(q + " " + f);
			
			ans = (ans + q * f) % mod;
			ans = (ans + ((q * b) % mod) * ((f * (f + 1) / 2) % mod)) % mod;
		}

		out.println(ans);
	}

	String nextLine() throws IOException {
		st = new StringTokenizer("");
		return in.readLine();
	}

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

	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}

	boolean EOF() throws IOException {
		while (!st.hasMoreTokens()) {
			String str = in.readLine();
			if (str == null)
				return true;
			st = new StringTokenizer(str);
		}
		return false;
	}

}
