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.math.BigInteger;
import java.util.*;

public class Main
{
    static long[][] map;

    public static long dfs(int[][] min,int[] arr,int from,int to)
    {
        if(from>=to)
            return 1;
        int pos=min[from][to];
        if(map[from][to]>=0)
        {
            return map[from][to];
        }
        long left=0,right=0,res=0;
        for(int i=from;i<=pos;i++)
        {
            left=(left+dfs(min, arr, from, i - 1) * dfs(min, arr, i, pos - 1))%998244353;
        }
        for(int i=pos;i<=to;i++)
        {
            right=(right+dfs(min, arr, pos + 1, i) * dfs(min, arr, i + 1, to))%998244353;
        }
        res=(left*right)%998244353;
        map[from][to]=res;
        return map[from][to];
    }

    public static void main(String[] args)
    {
        Scanner in=new Scanner(System.in);
        int m,n;
        int[] arr;
        int[][] min;
        n=in.nextInt();
        m=in.nextInt();
        arr=new int[m];
        min=new int[m][m];
        for(int i=0;i<m;i++)
            arr[i] = in.nextInt();
        map=new long[m][m];
        for(int i=0;i<m;i++)
            for(int j=0;j<m;j++)
                map[i][j]=-1;
        for(int i=0;i<m;i++)
            for(int j=i;j<m;j++)
            {
                if(i==j)
                    min[i][j]=i;
                else
                {
                    if(arr[j]<arr[min[i][j-1]])
                        min[i][j]=j;
                    else
                        min[i][j]=min[i][j-1];
                }
            }
        System.out.println(dfs(min,arr,0,arr.length-1));
    }
}
