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.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.SortedSet;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.TreeSet;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.InputStream;

public class _1178F1 {
	static int[] perm;
	static long[][] table;
	
	public static void main(String[] args) {
		InputReader in = new InputReader();
		int n = in.nextInt();
		int m = in.nextInt();
		perm = new int[n];
		table = new long[n+1][n+1];
		for (int i = 0; i < n; i++) {
			perm[i] = in.nextInt();
		}
		System.out.println(helper(0, n));
	}
	
	private static long helper(int a, int b) {
		if (b - a <= 1) {
			return 1;
		}
		long val = table[a][b];
		if (val != 0) {
			return val;
		}
		
		int min = 501;
		int minIndex = 501;
		for (int i = a; i < b; i++) {
			if (perm[i] < min) {
				min = perm[i];
				minIndex = i;
			}
		}
		long leftPos = 0; long rightPos = 0;
		for (int i = a; i <= minIndex; i++) {
			leftPos += helper(a, i) * helper(i, minIndex);
			leftPos %= 998244353;
		}
		for (int j = minIndex + 1; j <= b; j++) {
			rightPos += helper(minIndex + 1, j) * helper(j, b);
			rightPos %= 998244353;
		}
		long total = leftPos * rightPos % 998244353;
		table[a][b] = total;
		return total;
	}
	
	static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
 
        public InputReader() {
            reader = new BufferedReader(new InputStreamReader(System.in));
            tokenizer = null;
        }
 
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
 
        public int nextInt() {
            return Integer.parseInt(next());
        }
 
        public long nextLong() {
            return Long.parseLong(next());
        }
        
        public double nextDouble() {
        	return Double.parseDouble(next());
        }
        
        public String nextLine() {
        	String str = "";
        	try {
        		str = reader.readLine();
        	} catch (IOException e) {
        		e.printStackTrace();
        	}
        	return str;
        }
    }
}