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.Scanner;;

public class main {
	public static int min(int u, int v)
	{
		if ( u < v ) return u;
		else return v;
	}
	public static void main(String[] args) {
        // TODO Auto-generated method stub
		Scanner scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		int m = scanner.nextInt();
		long[][] f= new long[502][502];
		int[][] g= new int[502][502];
		int[] a= new int[502];
		final int MOD=998244353;
		long s=0;
		for(int i=1;i<=n;i++)
			a[i]=scanner.nextInt();
		for(int i=1;i<=n;i++)
		{
			int MIN=i;
			for(int j=i;j<=n;j++)
			{
				if ( a[j] < a[MIN] ) MIN=j;
				g[i][j]=MIN;
				//f[i][j]=1;
			}
			f[i][i]=1;
		}
		for(int i=n;i>0;i--)
			for(int j=i+1;j<=n;j++)
			{
				long u=f[i][g[i][j]-1]*2,v=f[g[i][j]+1][j]*2;
				for(int k=i;k<g[i][j];k++)
					u=(u+f[i][k]*f[k+1][g[i][j]-1])%MOD;
				for(int k=g[i][j]+1;k<j;k++)
					v=(v+f[g[i][j]+1][k]*f[k+1][j])%MOD;
				if ( g[i][j] == i ) u=1;
				if ( g[i][j] == j ) v=1;
				f[i][j]=(u*v)%MOD;
			}
//		for(int i=1;i<=n;i++)
//		{
//			for(int j=1;j<=n;j++)
//			{
//				System.out.print(f[i][j]+" ");
//			}
//			System.out.println(" ");
//		}
        System.out.println(f[1][n]);
    }
}