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.math.BigInteger;


public class Div2_272 {
	
	static public void main(String[] args) throws Exception{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] in=br.readLine().split(" ");
		long a=Long.parseLong(in[0]);
		long b=Long.parseLong(in[1]);
		long mod=1000000000+7;
		
		long num1=b*(b-1)/2;
		long num2=a*(a+1)/2;
		BigInteger big = new BigInteger(Long.toString(b)).multiply( new BigInteger(Long.toString(num1))).multiply( new BigInteger(Long.toString(num2)));
		BigInteger temp=  new BigInteger(Long.toString(a)).multiply( new BigInteger(Long.toString(num1)));
		
		big=big.add(temp);

		
		System.out.println(big.mod(new BigInteger("1000000007")));
	}
}