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.IOException;
import java.io.InputStream;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.InputMismatchException;
public class FGlobalRound4B {
	
	public static final long p = 998244353;

	public static int[] arr;
	public static int[][] indexLow;
	public static long[][] dp;
	
	public static void main(String[] args) {
		FastScanner in = new FastScanner(System.in);
		int N = in.nextInt();
		int M = in.nextInt();
		arr = new int[M];
		for(int i = 0; i < M; i++)
			arr[i] = in.nextInt() - 1;
		
		arr = compress(arr);
		
		long ans = ans(N, M, arr);
		System.out.println(ans);
	}
	
	public static int[] compress(int[] arr) {
		ArrayList<Integer> list = new ArrayList<>();
		list.add(arr[0]);
		for(int i = 1; i < arr.length; i++)
			if(list.get(list.size() - 1) != arr[i])
				list.add(arr[i]);
		
		int[] temp = new int[list.size()];
		for(int i = 0; i < temp.length; i++)
			temp[i] = list.get(i);
		
		return temp;
	}
	
	public static long ans(int N, int M, int[] arr) {
		if(arr.length > 2 * N)
			return 0;
		
		N = arr.length;
		
		@SuppressWarnings("unchecked")
		ArrayList<Integer>[] positions = new ArrayList[N];
		for(int i = 0; i < N; i++)
			positions[i] = new ArrayList<>();
		
		for(int i = 0; i < arr.length; i++) {
			positions[arr[i]].add(i);
		}
		
		indexLow = new int[N+1][N+1];
		for(int i = 0; i < indexLow.length; i++)
			Arrays.fill(indexLow[i], -1);
		for(int i = 0; i < N; i++) {
			indexLow[i][i+1] = i;
			int min = arr[i];
			for(int j = i + 2; j < N+1; j++) {
				if(arr[j-1] < min) {
					min = arr[j-1];
					indexLow[i][j] = j-1;
				}
				else {
					indexLow[i][j] = indexLow[i][j-1];
				}
			}
		}
		
		dp = new long[arr.length + 1][arr.length + 1];
		for(int i = 0; i < dp.length; i++)
			Arrays.fill(dp[i], -1);
		
		return get(0, arr.length, positions);
	}
	
	public static long get(int left, int right, ArrayList<Integer>[] positions) {
		if(right - left == 0)
			return 1;
		
		if(right - left == 1) {
			ArrayList<Integer> pos = positions[arr[left]];
			if(left <= pos.get(0) && pos.get(pos.size()-1) < right) {
				return 1;
			}
			return 0;
		}
		
		long temp = dp[left][right];
		if(temp != -1) {
			return temp; 
		}
		
		temp = 1;
		int indexMinColor = indexLow[left][right];
		int colorMin = arr[indexMinColor];
		ArrayList<Integer> pos = positions[colorMin];
		int index = Collections.binarySearch(pos, indexMinColor);
		
		if(index != 0 || pos.get(pos.size()-1) >= right) {
			return 0;
		}
		
		long leftAns = 0;
		for(int i = left; i <= indexMinColor; i++) {
			long t = (get(left, i, positions) * get(i, indexMinColor, positions)) % p;
			leftAns = (leftAns + t) % p;
		}
		
		temp = leftAns;
		while(index + 1 < pos.size() && pos.get(index+1) <= right) {
			long t = get(pos.get(index) + 1, pos.get(index+1), positions);
			temp = (temp * t) % p;
			index++;
		}
		
		long rightAns = 0;
		indexMinColor = pos.get(index);
		for(int i = indexMinColor + 1; i <= right; i++) {
			long t = (get(indexMinColor + 1, i, positions) * get(i, right, positions)) % p;
			rightAns = (rightAns + t) % p;
		}
		
		temp = (temp * rightAns) % p;
		
		return dp[left][right] = temp;
	}
	
	/**
	 * Source: Matt Fontaine
	 */
	static class FastScanner {
		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int chars;

		public FastScanner(InputStream stream) {
			this.stream = stream;
		}

		int read() {
			if (chars == -1)
				throw new InputMismatchException();
			if (curChar >= chars) {
				curChar = 0;
				try {
					chars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (chars <= 0)
					return -1;
			}
			return buf[curChar++];
		}

		boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		boolean isEndline(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}

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

		public long nextLong() {
			return Long.parseLong(next());
		}

		public double nextDouble() {
			return Double.parseDouble(next());
		}

		public String next() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		public String nextLine() {
			int c = read();
			while (isEndline(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndline(c));
			return res.toString();
		}
	}

}
/*
3 3
1 2 3
outputCopy
5
inputCopy
2 3
1 2 1
outputCopy
1
inputCopy
2 3
2 1 2
outputCopy
0
inputCopy
7 7
4 5 1 6 2 3 7
outputCopy
165
inputCopy
8 17
1 3 2 2 7 8 2 5 5 4 4 4 1 1 6 1 1
outputCopy
20
*/