This is the second subtask of problem F. The only differences between this and the first subtask are the constraints on the value of m and the time limit. It is sufficient to solve this subtask in order to hack it, but you need to solve both subtasks in order to hack the first 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 ≤ 10^6) — 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.

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


2 3
1 2 1


Output


1


Input


2 3
2 1 2


Output


0


Input


7 7
4 5 1 6 2 3 7


Output


165


Input


8 17
1 3 2 2 7 8 2 5 5 4 4 4 1 1 6 1 1


Output


20

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>

In the second example, Alice must first paint segment 0 3 with colour 1 and then segment 1 2 with colour 2. 
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);
        TaskF2 solver = new TaskF2();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskF2 {
        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);
                right[c[i]] = Math.max(right[c[i]], i);
            }

            if (m > 2 * n) {
                out.println(0);
                return;
            }

            int[][] minCol = 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];
                    if (l < r && minCol[l][r] > minCol[l + 1][r]) {
                        minCol[l][r] = minCol[l + 1][r];
                    }
                    int col = minCol[l][r];

                    if (left[col] >= l && right[col] <= r) {
                        long wl = left[col] == l ? 1 : f[l][left[col] - 1];
                        long wr = right[col] == r ? 1 : f[right[col] + 1][r];
                        long w = wl * wr % MOD;
                        int p = left[col];
                        for (int i = left[col] + 1; i <= right[col]; i++) {
                            if (c[i] != col) {
                                continue;
                            }
                            if (p + 1 <= i - 1) {
                                w = w * d[p + 1][i - 1] % MOD;
                            }
                            p = i;
                        }
                        d[l][r] = w;
                    }

                    // 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());
        }

    }
}

