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>
import java.lang.*;
import java.util.*;
import java.io.*;

public class Main {
    void solve() {
        int n = ni(), m = ni();
        long dp[][] = new long[m + 1][m + 1];
        int a[] = new int[m + 1];
        for (int i = 1; i <= m; i++) {
            a[i] = ni();
        }
        int mn[][] = new int[m + 1][m + 1];
        for (int i = 1; i <= m; i++) mn[i][i] = i;
        for (int sz = 2; sz <= m; sz++) {
            for (int i = 1; i <= m; i++) {
                int j = i + sz - 1;
                if (j > m) break;
                if (a[mn[i + 1][j]] < a[mn[i][j - 1]]) mn[i][j] = mn[i + 1][j];
                else mn[i][j] = mn[i][j - 1];
            }
        }
        for(int i=1;i<=m;i++) dp[i][i]=1;
        for (int sz = 2; sz <= m; sz++) {
            for (int i = 1; i <= m; i++) {
                int j = i + sz - 1;
                if (j > m) break;
                long left=0,right=0;
                    for (int k = i; k <=mn[i][j]; k++) {
                        left = add(left, mul(i <= k - 1 ? dp[i][k - 1] : 1, k<=mn[i][j]-1?dp[k][mn[i][j]-1]:1));
                    }

                    for (int k = j; k >= mn[i][j]; k--) {
                        right = add(right,mul(mn[i][j]+1<=k?dp[mn[i][j]+1][k]:1,k+1<=j?dp[k+1][j]:1));
                    }

                    dp[i][j]=mul(left,right);

            }
        }
        pw.println(dp[1][n]);


    }

    long mul(long a, long b) {
        a *= b;
        if (a >= M) a %= M;
        return a;
    }

    long add(long a, long b) {
        a+=b;
        if(a>=M) a-=M;
        return a;
    }



    long M = 998244353;


    // END
    PrintWriter pw;
    StringTokenizer st;
    BufferedReader br;

    void run() throws Exception {
        br = new BufferedReader(new InputStreamReader(System.in));
        pw = new PrintWriter(System.out);
        long s = System.currentTimeMillis();
        solve();
        pw.flush();
    }
    public static void main(String[] args) throws Exception {
        new Main().run();
    }

    String ns() {
        while (st == null || !st.hasMoreElements()) {

            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    String nextLine() throws Exception {
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            throw new Exception(e.toString());
        }
        return str;
    }

    int ni() {
        return Integer.parseInt(ns());
    }

    long nl() {
        return Long.parseLong(ns());
    }

    double nd() {
        return Double.parseDouble(ns());
    }
}