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.*;
import java.util.*;

public class Main{
    final boolean isFileIO = false; 
    BufferedReader in;
    PrintWriter out;
    StringTokenizer st = new StringTokenizer("");
    String delim = " ";

    public static void main(String[] args) throws IOException {
        Main m = new Main();
        m.initIO();
        m.solve();
        m.in.close();
        m.out.close();
    }

    public void initIO() throws IOException {
        if(!isFileIO) {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
        } else {
            in = new BufferedReader(new FileReader("14.in"));
            out = new PrintWriter("output.txt"); 
        }
    }
  
    String nextToken() throws IOException {
        if(!st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken(delim);
    }
 
    String readLine() throws IOException {
        return in.readLine();
    }

    int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    public void solve() throws IOException {
        long a, b;
        a = nextLong(); b = nextLong();

        long ans = 0;
        long mod = 1_000_000_007;
        long inv = 500000004;
        for(long i = 1; i < b; i++) {
            long nModI = i * a + 1;
            long nDivI = (nModI - 1) / i;            
            ans += ((((((2 * (i + b * i) + (((b * i) % mod) * (nDivI - 1)) % mod)) % mod) * nDivI) % mod) * inv) % mod;
            ans %= mod;

        }
        out.println(ans);
    }

}

class Utils {

    public static long binpow(long a, long exp, long mod) {
        if(exp == 0) {
            return 1;
        }
        if(exp % 2 == 0) {
            long temp = binpow(a, exp / 2, mod);
            return (temp * temp) % mod;
        } else {
            return (binpow(a, exp - 1, mod) * a) % mod;
        }
    }

    public static long inv(long a, long mod) {
        return binpow(a, mod - 2, mod);
    }

    public static long addmod(long a, long b, long mod) {
        return (a + b + 2 * mod) % mod;
    }
}
