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.util.*;

import static java.lang.Math.*;

public class F1 {
    static int n, m, a[];
    static int minValPos[][];
    static long dp[][], mod = 998244353;

    public static void main(String[] args) throws Exception {
        new Thread(null ,new Runnable(){
            public void run(){
                try{solveIt();} catch(Exception e){e.printStackTrace(); System.exit(1);}
            }
        },"Main",1<<28).start();
    }
    
    public static void solveIt() throws Exception {
        Scanner in = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        n = in.nextInt();
        m = in.nextInt();
        a = new int[n + 1];
        minValPos = new int[n + 1][n + 1];
        dp = new long[n + 1][n + 1];
        for (int i = 1; i <= n; i++) {
            a[i] = in.nextInt();
        }
        for (int i = 1; i <= n; i++) {
            int m = i;
            for (int j = i; j <= n; j++) {
                if (a[j] < a[m]) m = j;
                minValPos[i][j] = m;
            }
        }
        for (long a[] : dp) Arrays.fill(a, -1);
        pw.println(solve(1, n));
        pw.close();
    }

    static long solve(int l, int r) {
        if (l >= r) return 1;
        if (dp[l][r] != -1) return dp[l][r];
        int p = minValPos[l][r];
        long a = 0, b = 0;
        for (int i = l; i <= p; i++) {
            a += ((solve(l, i - 1) * solve(i, p - 1)) % mod);
            a %= mod;
        }
        for (int j = p; j <= r; j++) {
            b += ((solve(p + 1, j) * solve(j + 1, r)) % mod);
            b %= mod;
        }
        long res = (a * b) % mod;
        return dp[l][r] = res;
    }

    static void debug(Object... obj) {
        System.err.println(Arrays.deepToString(obj));
    }
}