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.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Queue;
import java.util.Set;
import java.util.StringTokenizer;

public class Code {
    public static void main(String[] args) {
        // Use the Scanner class
        MyScanner sc = new MyScanner();
        out = new PrintWriter(new BufferedOutputStream(System.out));

        long a = sc.nextLong();
        long b = sc.nextLong();

        long BASE = 1000000007L;
        long res = 0L;
        a = a % BASE;
        b = b % BASE;
        res = a%2 == 0 ? (a/2)*(a+1) : (a)*((a+1)/2);
        res = res % BASE;
        res *= b;
        res = res % BASE;
        res += a;
        res = res % BASE;

        long res3 = b%2 == 0 ? (b/2)*(b-1) : (b)*((b-1)/2);
        res3 = res3 %BASE;
        System.out.println((res*res3)%BASE);
    }


    static long f(int n, int L, int[] A) {
        long[] cur = new long[1+L];

        for (int i = 0; i <= L; ++i) {
            cur[i] = Long.MAX_VALUE;
        }
        cur[0] = 0;

        for (int j = 1; j <= n; ++j) {
            int vol = 1 << (j-1); // check this
            for (int i = 1; i <= L; ++i) {
                int del = Math.max(0, i - vol);
                cur[i] = Math.min(cur[i], cur[del] + A[j-1]);
            }
        }
        return cur[L];

    }

    public static PrintWriter out;

    //-----------MyScanner class for faster input----------
    public static class MyScanner {
       BufferedReader br;
       StringTokenizer st;

       public MyScanner() {
          br = new BufferedReader(new InputStreamReader(System.in));
       }

       String next() {
           while (st == null || !st.hasMoreElements()) {
               try {
                   st = new StringTokenizer(br.readLine());
               } catch (IOException e) {
                   e.printStackTrace();
               }
           }
           return st.nextToken();
       }

       int nextInt() {
           return Integer.parseInt(next());
       }

       long nextLong() {
           return Long.parseLong(next());
       }

       double nextDouble() {
           return Double.parseDouble(next());
       }

       String nextLine(){
           String str = "";
       try {
          str = br.readLine();
       } catch (IOException e) {
          e.printStackTrace();
       }
       return str;
       }

    }
}
