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}.
// Forza Juve
import java.io.*;
import java.util.*;
import java.awt.Point;

public class Template {

    BufferedReader in;
    PrintWriter out;
    StringTokenizer st;

    String next() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(in.readLine());
            } catch (Exception e) {
            }
        }
        return st.nextToken();
    }

    int nextInt() {
        return Integer.parseInt(next());
    }

    long nextLong() {
        return Long.parseLong(next());
    }

    double nextDouble() {
        return Double.parseDouble(next());
    }

    public void run() throws Exception {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.flush();
        out.close();
        in.close();
    }

    public void solve() throws Exception {
        long a = nextLong();
        long b = nextLong();

        long mod = 1000000007;

        long aa = (a * (a + 1) / 2) % mod;
        long bb = ((b - 1) * b / 2) % mod;
        long aabb = (aa * bb) % mod;
        long ans = (aabb * b) % mod;
        ans += (a * bb) % mod;
        ans %= mod;
        
        out.println(ans);
    }

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