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 C {
	
	static class Scanner{
		BufferedReader br=null;
		StringTokenizer tk=null;
		public Scanner(){
			br=new BufferedReader(new InputStreamReader(System.in));
		}
		public String next() throws IOException{
			while(tk==null || !tk.hasMoreTokens())
				tk=new StringTokenizer(br.readLine());
			return tk.nextToken();
		}
		public int nextInt() throws NumberFormatException, IOException{
			return Integer.valueOf(next());
		}
		public long nextLong() throws NumberFormatException, IOException{
			return Long.valueOf(next());
		}
		public double nextDouble() throws NumberFormatException, IOException{
			return Double.valueOf(next());
		}
	}
	
	static long mod = 1000000007;
	
	
	static long gauss(long n){
		return ((n*(n+1))/2) % mod;
	}
	
	public static void main(String args[]) throws NumberFormatException, IOException{
		Scanner sc = new Scanner();
		int a = sc.nextInt();
		int b = sc.nextInt();
		long ans = 0;
		long G = gauss(a);
		for(long i = 1; i < b; i++){
			long first = ( G * ((b * i)%mod)) % mod;
			long second = (a * i) % mod;
			ans = (ans + first + second) % mod;
		}
		System.out.println(ans);
	}

}
