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 PYTHON3):

Solution:
n, m = map(int, input().split())
l = list(map(int, input().split()))
index = [[0 for i in range(n)] for j in range(n)]
for i in range(n):
	mini = 10000000000000
	for j in range(i, n):
		if l[j] < mini:
			inde = j
			mini = l[j]
		index[i][j] = inde
prime = 998244353
d = {}
val = [[1 for i in range(n + 1)] for j in range(n + 1)]
for i in range(n):
	for j in range(n - i):
		if i == 0:
			val[j][j + i] = 1
		elif i == 1:
			val[j][j + i] = 2
		else:
			ind = index[j][j + i]
			sumap = 0
			sumak = 0
			for p in range(j, ind +1):
				sumap += (val[j][p - 1] * val[p][ind - 1]) % prime
			for k in range(ind, j + i + 1):
				sumak += (val[ind + 1][k] * val[k + 1][j + i]) % prime
			val[j][j + i] = (sumap * sumak) % prime
print(val[0][n-1])