Problem: This is the first subtask of problem F. The only differences between this and the second subtask are the constraints on the value of m and the time limit. You need to solve both subtasks in order to hack this one.

There are n+1 distinct colours in the universe, numbered 0 through n. There is a strip of paper m centimetres long initially painted with colour 0. 

Alice took a brush and painted the strip using the following process. For each i from 1 to n, in this order, she picks two integers 0 ≤ a_i < b_i ≤ m, such that the segment [a_i, b_i] is currently painted with a single colour, and repaints it with colour i. 

Alice chose the segments in such a way that each centimetre is now painted in some colour other than 0. Formally, the segment [i-1, i] is painted with colour c_i (c_i ≠ 0). Every colour other than 0 is visible on the strip.

Count the number of different pairs of sequences \\{a_i\}_{i=1}^n, \\{b_i\}_{i=1}^n that result in this configuration. 

Since this number may be large, output it modulo 998244353.

Input

The first line contains a two integers n, m (1 ≤ n ≤ 500, n = m) — the number of colours excluding the colour 0 and the length of the paper, respectively.

The second line contains m space separated integers c_1, c_2, …, c_m (1 ≤ c_i ≤ n) — the colour visible on the segment [i-1, i] after the process ends. It is guaranteed that for all j between 1 and n there is an index k such that c_k = j.

Note that since in this subtask n = m, this means that c is a permutation of integers 1 through n.

Output

Output a single integer — the number of ways Alice can perform the painting, modulo 998244353.

Examples

Input


3 3
1 2 3


Output


5


Input


7 7
4 5 1 6 2 3 7


Output


165

Note

In the first example, there are 5 ways, all depicted in the figure below. Here, 0 is white, 1 is red, 2 is green and 3 is blue.

<image>

Below is an example of a painting process that is not valid, as in the second step the segment 1 3 is not single colour, and thus may not be repainted with colour 2.

<image>

Solution (in JAVA):

Solution:
import java.io.*;
import java.nio.CharBuffer;
import java.util.NoSuchElementException;

public class P1178F1 {

    private static final int MOD = 998244353;

    public static void main(String[] args) {
        SimpleScanner scanner = new SimpleScanner(System.in);
        PrintWriter writer = new PrintWriter(System.out);

        int n = scanner.nextInt();
        int m = scanner.nextInt();
        int[] c = new int[n];
        for (int i = 0; i < n; ++i)
            c[i] = scanner.nextInt();
        long[][] f = new long[n][n];
        writer.println(solve(c, f, 0, n - 1));

        writer.close();
    }

    private static long solve(int[] c, long[][] f, int l, int r) {
        if (l >= r)
            return 1;
        if (f[l][r] > 0)
            return f[l][r];
        int min = Integer.MAX_VALUE;
        int mid = -1;
        for (int i = l; i <= r; ++i) {
            if (c[i] < min) {
                min = c[i];
                mid = i;
            }
        }
        long left = 0;
        for (int a = l; a <= mid; ++a) {
            left = (left + solve(c, f, l, a - 1) * solve(c, f, a, mid - 1) % MOD) % MOD;
        }
        long right = 0;
        for (int b = r; b >= mid; --b) {
            right = (right + solve(c, f, mid + 1, b) * solve(c, f, b + 1, r) % MOD) % MOD;
        }
        f[l][r] = left * right % MOD;
        return f[l][r];
    }


    private static class SimpleScanner {

        private static final int BUFFER_SIZE = 10240;

        private Readable in;
        private CharBuffer buffer;
        private boolean eof;

        SimpleScanner(InputStream in) {
            this.in = new BufferedReader(new InputStreamReader(in));
            buffer = CharBuffer.allocate(BUFFER_SIZE);
            buffer.limit(0);
            eof = false;
        }


        private char read() {
            if (!buffer.hasRemaining()) {
                buffer.clear();
                int n;
                try {
                    n = in.read(buffer);
                } catch (IOException e) {
                    n = -1;
                }
                if (n <= 0) {
                    eof = true;
                    return '\0';
                }
                buffer.flip();
            }
            return buffer.get();
        }

        void checkEof() {
            if (eof)
                throw new NoSuchElementException();
        }

        char nextChar() {
            checkEof();
            char b = read();
            checkEof();
            return b;
        }

        String next() {
            char b;
            do {
                b = read();
                checkEof();
            } while (Character.isWhitespace(b));
            StringBuilder sb = new StringBuilder();
            do {
                sb.append(b);
                b = read();
            } while (!eof && !Character.isWhitespace(b));
            return sb.toString();
        }

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

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

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