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.io.*;
import java.util.*;

public class F{
    static DataInputStream din;

    // Read one number, possibly negative, that fits into an int.
    // Ignores whitespace.
    private static int ri() throws IOException {
        int c = din.read();
        boolean negative = false;
        while (!('0' <= c && c <= '9')) {
            negative = c == '-';
            c = din.read();
        }
        int ret = 0;
        while ('0' <= c && c <= '9') {
            ret = 10 * ret + c - '0';
            c = din.read();
        }
        return ret * (negative ? -1 : 1);
    }
    static long N = 998244353;
    static long[][] sol;
    static long solve(int l, int r) {
        // Step 1. Find the lowest number. For each substr, get ceratin # of divisors.
        if (r - l <= 0) return 1;
        if (sol[l][r] != 0) return sol[l][r];
        int ind = l;
        for (int i = l+1; i <= r; i++) {
            if (c[i] < c[ind]) ind = i;
        }
        long tmp1 = 0, tmp2 = 0;
        // Step 2. Enumerate over all substrs.
        for (int i = l; i <= ind; i++) {
            long tmp = solve(l, i-1);
            tmp = tmp * solve(i, ind - 1) % N;
            tmp1 = (tmp1 + tmp) % N;
        }
        for (int j = ind; j <= r; j++) {
            long tmp = solve(ind + 1, j);
            tmp = tmp * solve(j + 1, r) % N;
            tmp2 = (tmp2 + tmp) % N;
        }

        sol[l][r] = (tmp1*tmp2) % N;
        return sol[l][r];
    }
    static int[] c;
    public static void main(String[] args) throws IOException{
        din = new DataInputStream(System.in);
        int n = ri(), m = ri();
        c = new int[n];
        for (int i = 0; i < n; i++) c[i] = ri();
        sol = new long[n][n];
        System.out.println(solve(0, n-1));
//        System.out.println(Arrays.deepToString(sol));
    }
}