Snuke has N dogs and M monkeys. He wants them to line up in a row.

As a Japanese saying goes, these dogs and monkeys are on bad terms. ("ken'en no naka", literally "the relationship of dogs and monkeys", means a relationship of mutual hatred.) Snuke is trying to reconsile them, by arranging the animals so that there are neither two adjacent dogs nor two adjacent monkeys.

How many such arrangements there are? Find the count modulo 10^9+7 (since animals cannot understand numbers larger than that). Here, dogs and monkeys are both distinguishable. Also, two arrangements that result from reversing each other are distinguished.

Constraints

* 1 ≤ N,M ≤ 10^5

Input

Input is given from Standard Input in the following format:


N M


Output

Print the number of possible arrangements, modulo 10^9+7.

Examples

Input

2 2


Output

8


Input

3 2


Output

12


Input

1 8


Output

0


Input

100000 100000


Output

530123477
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.IOException;
import java.util.NoSuchElementException;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FasterScanner in = new FasterScanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC {
        public void solve(int testNumber, FasterScanner in, PrintWriter out) {
            final long n = in.nextLong();
            final long m = in.nextLong();

            final long lim = 1000000007;

            final long d = Math.abs(n - m);

            long ans = 1;
            if (d >= 2) {
                ans = 0;
            } else {
                for (int i = 1; i <= Math.min(n, m); i++) {
                    ans *= i;
                    if (ans >= lim) ans %= lim;
                }
                ans *= ans;
                ans %= lim;
                if (n == m)
                    ans *= 2;
                else
                    ans *= Math.max(n, m);

                ans %= lim;
            }

            out.println(ans);
        }

    }

    static class FasterScanner {
        private InputStream in;
        private byte[] buffer;
        private int bufPointer;
        private int bufLength;

        public FasterScanner(InputStream in) {
            this.in = in;
            buffer = new byte[1024];
        }

        private boolean hasNextByte() {
            if (bufPointer < bufLength) return true;
            bufPointer = 0;
            try {
                bufLength = in.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            return (bufLength > 0);
        }

        private int readByte() {
            if (hasNextByte()) return buffer[bufPointer++];
            return -1;
        }

        private static boolean isPrintableChar(int c) {
            return (c >= 33 && c <= 126);
        }

        public boolean hasNext() {
            while (hasNextByte() && !isPrintableChar(buffer[bufPointer])) bufPointer++;
            return hasNextByte();
        }

        public long nextLong() {
            if (!hasNext()) throw new NoSuchElementException();
            long n = 0;
            boolean minus = false;
            int b = readByte();
            if (b == '-') {
                minus = true;
                b = readByte();
            }
            if (b < '0' || b > '9') throw new NumberFormatException();

            while (true) {
                if (b >= '0' && b <= '9') {
                    n *= 10;
                    n += b - '0';
                } else if (b == -1 || !isPrintableChar(b)) {
                    return minus ? -n : n;
                } else {
                    throw new NumberFormatException();
                }
                b = readByte();
            }
        }

    }
}

