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.*;
import java.util.*;
import java.math.BigInteger;

@SuppressWarnings("unused")
public class Main {
    public static void main(String[] args) throws FileNotFoundException {
        InputStream inputStream = System.in;
        // InputStream inputStream = new FileInputStream(new File(
        // "C:\\Users\\Administrator\\Codeblocks\\First\\1.in"));
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskA solver = new TaskA();
        solver.solve(1, in, out);
        out.close();
    }
}

// 使用对象时,首先考虑是否要采用深复制
class TaskA {
    static final int mod = (int) 1e9 + 7;
    static final int INF = 0x7fffffff;
    static final double eps = 1e-7;
    static final int maxn = (int) 1e6;

    // 声明全局变量

    public void solve(int testNumber, InputReader in, PrintWriter out) {
        int a = in.nextInt();
        int b = in.nextInt();

        long ans = 0;
        for (int k = 1; k <= a; k++) {
            long Q = (long) (b - 1) * (long) b / 2 % mod;
            long P = Q * b % mod;
            ans = (ans + (Q + P * k) % mod) % mod;
        }
        out.println(ans);
    }

}

class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public InputReader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream), 32768);
        tokenizer = null;
    }

    public String next() {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }

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

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

}
