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 static java.lang.Math.*;
import static java.lang.System.currentTimeMillis;
import static java.lang.System.exit;
import static java.lang.System.arraycopy;
import static java.util.Arrays.sort;
import static java.util.Arrays.binarySearch;
import static java.util.Arrays.fill;
import java.util.*;
import java.io.*;

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));
        out = new PrintWriter(System.out);
        solve();
        in.close();
        out.close();
    }
    
    final long MOD = 1000L * 1000L * 1000L + 7L;
    
    private void solve() throws IOException {
        
        long a = nextLong();
        long b = nextLong();
        int n = (int)b;
        long ans = 0;
        
        for (int i = 1; i < n; i++) {
            long cur = a * (a + 1) / 2;
            cur %= MOD;
            cur *= (b * (long)i) % MOD;
            cur %= MOD;
            cur += (a * (long)i) % MOD;
            cur %= MOD;
            ans += cur;
            ans %= MOD;
        }
        out.println(ans);
        
    }
    
    void chk(boolean b) {
        if (b)
            return;
        System.out.println(new Error().getStackTrace()[1]);
        exit(999);
    }
    void deb(String fmt, Object... args) {
        System.out.printf(Locale.US, fmt + "%n", args);
    }
    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());
    }
    String nextLine() throws IOException {
        st = new StringTokenizer("");
        return in.readLine();
    }
    boolean EOF() throws IOException {
        while (!st.hasMoreTokens()) {
            String s = in.readLine();
            if (s == null)
                return true;
            st = new StringTokenizer(s);
        }
        return false;
    }
}
