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


public class A {

    static BufferedReader br;
    static StringTokenizer st;
    public static void main(String[] args) throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        long a = nextLong();
        long b = nextLong();
        int mod = (int) (1e9+7);
        long ans = 0;
        for (int q = 1; q < b; q++) {
            long s = a*(a+1)/2 % mod * b % mod;
            s += a;
            if (s >= mod)
                s -= mod;
            s = s * q % mod;
            ans += s;
            if (ans >= mod)
                ans -= mod;
        }
        System.out.println(ans);
    }
    
    private static int nextInt() throws IOException {
        return Integer.parseInt(next());
    }
    private static long nextLong() throws IOException {
        return Long.parseLong(next());
    }
    private static double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }
    private static String next() throws IOException {
        while (st==null || !st.hasMoreTokens())
            st = new StringTokenizer(br.readLine());
        return st.nextToken();
    }

}
