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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastScanner in = new FastScanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskF1 solver = new TaskF1();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskF1 {
        public void solve(int testNumber, FastScanner in, PrintWriter out) {
            final int MOD = 998244353;
            int n = in.nextInt();
            int m = in.nextInt();
            int[] c = new int[m];
            int last = -1;
            int len = 0;
            for (int i = 0; i < m; i++) {
                int x = in.nextInt() - 1;
                if (x == last) {
                    continue;
                }
                c[len++] = x;
                last = x;
            }
            m = len;
            c = Arrays.copyOf(c, m);
            int[] left = new int[n];
            int[] right = new int[n];
            Arrays.fill(left, m - 1);
            Arrays.fill(right, 0);
            for (int i = 0; i < m; i++) {
                left[c[i]] = Math.min(left[c[i]], i - 1);
                right[c[i]] = Math.max(right[c[i]], i);
            }

            if (m > 2 * n) {
                out.println(0);
                return;
            }
            int[][] minCol = new int[m][m];
            int[][] minColPos = new int[m][m];
            // Number of ways to finish the colouring of the segment [l, r]
            // which is currently monochromatic.
            long[][] d = new long[m][m];
            // Number of ways to finish the colouring of the segment [l, r]
            // which consists of at most two monochromatic segments.
            long[][] f = new long[m][m];
            for (int l = m - 1; l >= 0; l--) {
                for (int r = l; r < m; r++) {
                    minCol[l][r] = c[l];
                    minColPos[l][r] = l;
                    if (l < r && minCol[l][r] > minCol[l + 1][r]) {
                        minCol[l][r] = minCol[l + 1][r];
                        minColPos[l][r] = minColPos[l + 1][r];
                    }
                    int j = minColPos[l][r];

                    long wl = j == l ? 1 : f[l][j - 1];
                    long wr = j == r ? 1 : f[j + 1][r];
                    d[l][r] = wl * wr % MOD;

                    // All col0 or all col1.
                    f[l][r] = 2 * d[l][r] % MOD;
                    // Col0 ends at a.
                    for (int a = l; a < r; a++) {
                        f[l][r] = (f[l][r] + d[l][a] * d[a + 1][r]) % MOD;
                    }
                }
            }
            out.println(d[0][m - 1]);
        }

    }

    static class FastScanner {
        private BufferedReader in;
        private StringTokenizer st;

        public FastScanner(InputStream stream) {
            in = new BufferedReader(new InputStreamReader(stream));
        }

        public String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    String rl = in.readLine();
                    if (rl == null) {
                        return null;
                    }
                    st = new StringTokenizer(rl);
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }
}