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.*;
import java.text.DecimalFormat;


public class F { 

	static class Reader 
    { 
        final private int BUFFER_SIZE = 1 << 16; 
        private DataInputStream din; 
        private byte[] buffer; 
        private int bufferPointer, bytesRead; 
  
        public Reader() 
        { 
            din = new DataInputStream(System.in); 
            buffer = new byte[BUFFER_SIZE]; 
            bufferPointer = bytesRead = 0; 
        } 
  
        public Reader(String file_name) throws IOException 
        { 
            din = new DataInputStream(new FileInputStream(file_name)); 
            buffer = new byte[BUFFER_SIZE]; 
            bufferPointer = bytesRead = 0; 
        } 
  
        public String readLine() throws IOException 
        { 
            byte[] buf = new byte[1000000]; // line length 
            int cnt = 0, c; 
            while ((c = read()) != -1) 
            { 
                if (c == '\n') 
                    break; 
                buf[cnt++] = (byte) c; 
            } 
            return new String(buf, 0, cnt); 
        } 
  
        public int nextInt() throws IOException 
        { 
            int ret = 0; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
            do
            { 
                ret = ret * 10 + c - '0'; 
            }  while ((c = read()) >= '0' && c <= '9'); 
  
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        public long nextLong() throws IOException 
        { 
            long ret = 0; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
            do { 
                ret = ret * 10 + c - '0'; 
            } 
            while ((c = read()) >= '0' && c <= '9'); 
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        public double nextDouble() throws IOException 
        { 
            double ret = 0, div = 1; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
  
            do { 
                ret = ret * 10 + c - '0'; 
            } 
            while ((c = read()) >= '0' && c <= '9'); 
  
            if (c == '.') 
            { 
                while ((c = read()) >= '0' && c <= '9') 
                { 
                    ret += (c - '0') / (div *= 10); 
                } 
            } 
  
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        private void fillBuffer() throws IOException 
        { 
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE); 
            if (bytesRead == -1) 
                buffer[0] = -1; 
        } 
  
        private byte read() throws IOException 
        { 
            if (bufferPointer == bytesRead) 
                fillBuffer(); 
            return buffer[bufferPointer++]; 
        } 
  
        public void close() throws IOException 
        { 
            if (din == null) 
                return; 
            din.close(); 
        } 
    } 

    static void oracle(int graph[][], int dist[][], int V) 
    { 
        int i, j, k; 
  
        for (i = 1; i < V; i++) 
            for (j = 1; j < V; j++) 
                dist[i][j] = graph[i][j]; 
  
        for (k = 1; k < V; k++) 
        { 
            for (i = 1; i < V; i++) 
            { 
                for (j = 1; j < V; j++) 
                { 
                    if (dist[i][k] + dist[k][j] < dist[i][j]) 
                        dist[i][j] = dist[i][k] + dist[k][j]; 
                } 
            } 
        } 
    } 

    static long v[][];

    static long find(long c[], int l, int r){
        if(l>=r)
            return 1;
        if(v[l][r]>0)
            return v[l][r];
        int i;
        int mini=l;
        long min=c[l];
        for(i=l+1;i<=r;i++){
            if(c[i]<min){
                min=c[i];
                mini=i;
            }
        }
        int mod=998244353;
        long sum1=0,sum2=0,p=1;
        for(i=l-1;i<mini;i++){
            sum1+=find(c,l,i)*find(c,i+1,mini-1)%mod;
            sum1%=mod;
        }
        for(i=mini;i<=r;i++){
            sum2+=find(c,mini+1,i)*find(c,i+1,r)%mod;
            sum2%=mod;
        }
        v[l][r]=(sum1*sum2)%mod;
        return v[l][r];
    }

    public static void main(String args[])throws IOException{
    	Reader sc=new Reader(); 
        // Scanner sc=new Scanner(System.in);
        PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        int i,j;
        int n=sc.nextInt();
        int m=sc.nextInt();
        v=new long[n+1][n+1];
        long c[]=new long[n];
        for(i=0;i<n;i++){
            c[i]=sc.nextInt();
        }
        int mod=998244353;
        long val=find(c,0,n-1);
        pw.println(val);
        pw.close();        
    }
}