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.InputStreamReader;
import java.util.StringTokenizer;


public class A {
	static long MOD = 1000000007;
	public static void main(String[] args) throws Exception{
		long a = readLong();
		long b = readLong();
		long ans = 0;
		
		long bsum = (b*(b-1))/2;
		long asum = (a*(a+1))/2;
		bsum%=MOD;
		asum%=MOD;
		
		long basum = b*asum;
		basum%=MOD;
		basum *= bsum;
		basum%=MOD;
		
		ans = a*bsum + basum;
		ans%=MOD;
		
		System.out.println(ans);
	}

	static BufferedReader stdin = new BufferedReader(new InputStreamReader(System.in));
	static StringTokenizer st = new StringTokenizer(" ");
	static String readString() throws Exception{
		while(!st.hasMoreTokens()){
			st = new StringTokenizer(stdin.readLine());
		}
		return st.nextToken();
	}
	static int readInt() throws Exception {
		return Integer.parseInt(readString());
	}
	static long readLong() throws Exception {
		return Long.parseLong(readString());
	}
}
