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:
/*
 * @author derrick20
 */
import java.io.*;
import java.util.*;

public class colorStripREDO {
    public static void main(String args[]) throws Exception {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

        N = sc.nextInt();
        int M = sc.nextInt();

        colors = new int[N];
        for (int i = 0; i < colors.length; i++) {
            colors[i] = sc.nextInt();
        }

        // from bound i to j, what is the position of lowest valued index
        // notice that it doesn't matter WHAT the value is, just that
        // it is the lowest
        indexOfMin = new int[N][N];
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                // literally search that i -> j range for the lowest value
                // this isn't too bad, N^3 is fine for 500
                int index = i; // just let it be the first in the range arbitrarily
                for (int k = i; k <= j; k++) {
                    if (colors[k] < colors[index]) {
                        index = k;
                    }
                }
                indexOfMin[i][j] = index;
            }
        }

        memo = new long[N][N];

        for (int i = 0; i < N; i++) {
            // length 1 base case
            memo[i][i] = 1;
            // length 2 base case
            if (i + 1 < N) {
                memo[i][i + 1] = 2;
            }
        }

        out.println(solve(0, N - 1));

        out.close();
    }

    static long[][] memo;
    static int[][] indexOfMin;
    static int[] colors;
    static int N;
    static long mod = 998244353;

    static long solve(int i, int j) {
        // we should never have any out of bounds issues
        if (memo[i][j] != 0) {
            return memo[i][j];
        }
        else {
            long left = 0;
            long right = 0;
            int pivot = indexOfMin[i][j];

            if (i != pivot) {
                // keep picking the left bound to which we will paint to
                for (int l = i + 1; l <= pivot - 1; l++) {
                    left += (solve(i, l - 1) * solve(l, pivot - 1)) % mod;
                    left %= mod;
                }
                // we have to add the final case of 0:len as the segment because it
                // isn't easy to fit that into the for loop without a bunch of edge cases
                left += solve(i, pivot - 1);
                left %= mod;
                left += solve(i, pivot - 1);
                left %= mod;
            }
            else {
                left = 1;
            }

            if (j != pivot) {
                // keep picking the right bound to paint to, independent of the left bound
                for (int r = pivot + 1; r <= j - 1; r++) {
                    right += (solve(pivot + 1, r) * solve(r + 1, j)) % mod;
                    right %= mod;
                }
                // add the len:0 case
                right += solve(pivot + 1, j);
                right %= mod;
                // As well as the 0:len case
                right += solve(pivot + 1, j);
                right %= mod;
            }
            else {
                right = 1;
            }

            return memo[i][j] = (left * right) % mod;
        }
    }

    static class Scanner {
        BufferedReader br;
        StringTokenizer st;

        Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        Scanner(FileReader s) {
            br = new BufferedReader(s);
        }

        String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

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

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