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.util.*;
import java.io.*;
import java.lang.*;
import java.math.*;
public class F {
    public static void main(String[] args) throws Exception {
        BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        // int n = Integer.parseInt(bf.readLine());
        StringTokenizer st = new StringTokenizer(bf.readLine());

        int n = Integer.parseInt(st.nextToken());
        int m = Integer.parseInt(st.nextToken());
        st = new StringTokenizer(bf.readLine());
        int[] a = new int[n]; for(int i=0; i<n; i++) a[i] = Integer.parseInt(st.nextToken());
        int mod = 998244353;

        long[][] dp = new long[n+1][n+1];
        for(int length = 0; length<=n; length++) {
          for(int start = 0; start+length<=n; start++) {
            if(length==1) dp[start][length] = 1;
            else if(length == 0) dp[start][length] = 1;
            else {
              int min = 1000001;
              int index = -1;
              for(int i=start; i<start+length; i++) {
                if(a[i] < min) {
                  min = a[i];
                  index = i;
                }
              }
              long val1 = 0;
              for(int i=0; i<=index-start; i++) {
                val1 += 1L * dp[start][i] * dp[start+i][index-start-i] % mod;
                val1 %= mod;
              }
              long val2 = 0;
              for(int i=0; i<=start+length-index-1; i++) {
                val2 += 1L * dp[index+1][i] * dp[index+1+i][start+length-index-1-i] % mod;
                val2 %= mod;
              }
              dp[start][length] = 1L*val1*val2 % mod;
            }
          }
        }
        out.println((dp[0][n] % mod));


        out.close(); System.exit(0);
    }
}
