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.util.*;
import java.io.*;

public class F14 {
    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        int n = sc.nextInt(); int m = sc.nextInt();
        long mod = 998244353;
        int [] a = new int[m];
        for (int i = 0; i < m; i++) a[i] = sc.nextInt();
        long [][] dp = new long[m][m];
        for (int len = 1; len <= m; len++) {
            for (int i = 0; i < m - len + 1; i++) {
                int j = i + len - 1;
                if (len == 1) {
                    dp[i][j] = 1;
                    continue;
                }
                int minIndex = -1; int min = Integer.MAX_VALUE;
                for (int k = i; k <= j; k++) {
                    if (a[k] < min) {
                        min = a[k]; minIndex = k;
                    }
                }
                long leftSum = 0;
                for (int left = minIndex; left >= i; left--) {
                    long r = 1; long l = 1;
                    if (left < minIndex) r = dp[left][minIndex - 1];
                    if (left > i) l = dp[i][left - 1];
                    leftSum = (leftSum + (r * l) % mod) % mod;
                }
                long rightSum = 0;
                for (int right = minIndex; right <= j; right++) {
                    long l = 1; long r = 1;
                    if (right > minIndex) l = dp[minIndex + 1][right];
                    if (right < j) r = dp[right + 1][j];
                    rightSum = (rightSum + (l * r) % mod) % mod;
                }
                dp[i][j] = (leftSum * rightSum) % mod;
            }
        }
        out.println(dp[0][m - 1]);
        out.close();
    }


    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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

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

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }


    }

}