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.awt.*;
import java.io.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.PrintStream;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
import java.util.Arrays;
import java.util.TreeMap;

public class CF_1178F1 {
    static int n, m;
    static int[] c;

    static long[][] dp;

    static int MOD = 998244353;

    public static void main(String[] args) {
        Scanner in = new Scanner(new BufferedInputStream(System.in));
        PrintStream out = new PrintStream(new BufferedOutputStream(System.out));

        n = in.nextInt();
        m = in.nextInt();
        c = new int[n];

        dp = new long[n][];
        for(int i = 0; i < n; i++){
            dp[i] = new long[n];
            Arrays.fill(dp[i], -1);
        }

        for (int i = 0; i < n; i++) c[i] = in.nextInt();

        out.println(solve(0, n - 1));

        out.close();
        in.close();
    }

    static long solve(int l, int r) {
        if (l == r) return 1;

        if (dp[l][r] != -1) return dp[l][r];

        long curr = Integer.MAX_VALUE;
        int pos = -1;
        for (int i = l; i <= r; i++) {
            if (c[i] < curr) {
                curr = c[i];
                pos = i;
            }
        }

        long a = 0;
        long b = 0;
        for (int l2 = l; l2 <= pos; l2++) {
            long add = 1;

            if (l2 != l) add = (add * solve(l, l2 - 1)) % MOD;
            if (l2 != pos) add = (add * solve(l2, pos - 1)) % MOD;

            a = (a + add) % MOD;
        }

        for (int r2 = pos; r2 <= r; r2++) {
            long add = 1;
            if (r2 != pos) add = (add * solve(pos + 1, r2));
            if (r2 != r) add = (add * solve(r2 + 1, r)) % MOD;

            b = (b + add) % MOD;
        }
        long ret = (a * b) % MOD;

        dp[l][r] = ret;
        return ret;
    }
}