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.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author anand.oza
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        F2LongColorfulStrip solver = new F2LongColorfulStrip();
        solver.solve(1, in, out);
        out.close();
    }

    static class F2LongColorfulStrip {
        static final NumberTheory.ModM m = new NumberTheory.ModM(998244353);
        int length;
        int[] count;
        int[] left;
        int[] right;
        int[] c;

        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int colors = in.nextInt();
            length = in.nextInt();

            int[] cWithDups = new int[length];
            for (int i = 0; i < length; i++) {
                cWithDups[i] = in.nextInt() - 1;
            }

            c = new int[length];
            c[0] = cWithDups[0];
            int index = 1;
            for (int i = 1; i < length; i++) {
                if (cWithDups[i] == cWithDups[i - 1])
                    continue;
                c[index++] = cWithDups[i];
            }
            int[] compressedC = new int[index];
            System.arraycopy(c, 0, compressedC, 0, index);
            c = compressedC;
            length = c.length;

            //System.out.println(Arrays.toString(c));

            left = new int[colors];
            Arrays.fill(left, Integer.MAX_VALUE);
            right = new int[colors];
            Arrays.fill(right, Integer.MIN_VALUE);
            count = new int[colors];
            for (int i = 0; i < length; i++) {
                count[c[i]]++;
                left[c[i]] = Math.min(left[c[i]], i);
                right[c[i]] = Math.max(right[c[i]], i);
            }

            //System.out.println(Arrays.toString(left));
            //System.out.println(Arrays.toString(right));

            long answer = solve(0, length - 1);

            out.println(answer);
        }

        private long solve(int leftIndex, int rightIndex) {
            //System.out.println("L-R : " + leftIndex + " " + rightIndex);
            long answer = 1;
            List<Integer> currentRun = new ArrayList<>();
            for (int i = leftIndex; i <= rightIndex; i++) {
                int curColor = c[i];
                currentRun.add(curColor);
                if (count[curColor] != 1) {
                    int left = i + 1;
                    i++;
                    for (; i <= right[curColor]; i++) {
                        if (c[i] < curColor) {
                            return 0;
                        }
                        if (c[i] == curColor) {
                            long sub = solve(left, i - 1);
                            answer = m.mult(answer, sub);
                            left = i + 1;
                        }
                    }
                    i--;
                }
            }
            //System.out.println("CR: " + currentRun);
            answer = m.mult(answer, solveF1(array(currentRun)));

            return answer;
        }

        private static int[] array(List<Integer> l) {
            int[] a = new int[l.size()];
            for (int i = 0; i < l.size(); i++) {
                a[i] = l.get(i);
            }
            return a;
        }

        private long solveF1(int[] c) {
            int n = c.length;
            long[][] dp = new long[n][n]; // [i][j] is interval [i, j]
            for (int i = 0; i < n; i++)
                for (int j = 0; j < n; j++)
                    dp[i][j] = -1;
            for (int i = 0; i < n; i++) {
                dp[i][i] = 1;
            }

            return compute(0, n - 1, dp, c, n);
        }

        private long compute(int left, int right, long[][] dp, int[] c, int n) {
            if (left < 0 || right >= n || left > right)
                return 1;

            if (dp[left][right] >= 0)
                return dp[left][right];

            int min = left;
            for (int i = left; i <= right; i++) {
                if (c[i] < c[min]) {
                    min = i;
                }
            }

            long leftResult = 0;
            long rightResult = 0;
            for (int l = left; l <= min; l++) {
                long sub = 1;
                sub = m.mult(sub, compute(left, l - 1, dp, c, n));
                sub = m.mult(sub, compute(l, min - 1, dp, c, n));
                leftResult = m.add(leftResult, sub);
            }
            for (int r = min; r <= right; r++) {
                long sub = 1;
                sub = m.mult(sub, compute(min + 1, r, dp, c, n));
                sub = m.mult(sub, compute(r + 1, right, dp, c, n));
                rightResult = m.add(rightResult, sub);
            }

            long result = m.mult(leftResult, rightResult);

            dp[left][right] = result;

            //        System.out.format("%5d ", result);
            //        System.out.format("%5d (%4d)    ", min, c[min]);
            //        for (int i = left; i <= right; i++) {
            //            System.out.format("%d, ", c[i]);
            //        }
            //        //System.out.println();

            return result;
        }

    }

    static class NumberTheory {
        public abstract static class Modulus<M extends NumberTheory.Modulus<M>> {
            ArrayList<Long> factorial = new ArrayList<>();
            ArrayList<Long> invFactorial = new ArrayList<>();

            public abstract long modulus();

            public Modulus() {
                super();
                factorial.add(1L);
                invFactorial.add(1L);
            }

            public long normalize(long x) {
                x %= modulus();
                if (x < 0)
                    x += modulus();
                return x;
            }

            public long add(long a, long b) {
                return normalize(a + b);
            }

            public long mult(long a, long b) {
                return normalize(a * b);
            }

        }

        public static class ModM extends NumberTheory.Modulus<NumberTheory.ModM> {
            private final long modulus;

            public ModM(long modulus) {
                this.modulus = modulus;
            }

            public long modulus() {
                return modulus;
            }

        }

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

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

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

    }
}

