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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;


public class Main {

    
    public static void main(String[] args) throws IOException {
        
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = new BufferedWriter(new PrintWriter(System.out));
        
        String[] ln1 = br.readLine().split(" ");
        
        int n = Integer.parseInt(ln1[0]);
        
        int m = Integer.parseInt(ln1[1]);
        
        BigInteger a = new BigInteger(String.valueOf(n));
        
        BigInteger b = new BigInteger(String.valueOf(m));
        
        BigInteger c = new BigInteger(String.valueOf(m-1));
        
        BigInteger d = new BigInteger(String.valueOf(n+1));
        
        BigInteger TWO = new BigInteger(String.valueOf(2));
        
        BigInteger e = (b.multiply(c)).divide(TWO);
        //System.out.println(a.toString()+" "+c.toString()+" "+e.toString());
        
        BigInteger f = (b.multiply(a).multiply(d)).divide(TWO);
        
        BigInteger g = f.add(a);
        
        BigInteger h = g.multiply(e);
        
        BigInteger mod = new BigInteger(String.valueOf("1000000007"));
        
        h=h.mod(mod);
        
        bw.append(h.toString()+"\n");
        
        
        bw.close();
    }

    private static int solve(int d, int n) {
        
        int y = n-d;
        int x = d-y;
        
        if(x>=0 && y>=0)    return x+y;
        else return -1;
    }
}
