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>
class SparseTable():
    """区間取得クエリをO(1)で答えるデータ構造をO(NlogN)で構築する
    query(l, r): 区間[l, r)に対するクエリに答える
    """
    def __init__(self, array, n):
        n = len(array)
        self.row_size = n.bit_length()

        # log_tableを構築する
        # log_table = [0, 0, 1, 1, 2, 2, 2, 2, ...]
        self.log_table = [0] * (n + 1)
        for i in range(2, n + 1):
            self.log_table[i] = self.log_table[i//2] + 1

        # sparse_tableを構築する
        self.sparse_table = [[0] * n for _ in range(self.row_size)]
        for i in range(n):
            self.sparse_table[0][i] = array[i]
        for row in range(1, self.row_size):
            for i in range(n - (1 << row) + 1):
                self.sparse_table[row][i] = self._merge(self.sparse_table[row - 1][i], \
                                            self.sparse_table[row - 1][i + (1 << row - 1)])

    def _merge(self, num1, num2):
        """クエリの内容"""
        return min(num1, num2)

    def query(self, l, r):
        """区間[l, r)に対するクエリに答える"""
        row = self.log_table[r - l]
        return self._merge(self.sparse_table[row][l], self.sparse_table[row][r - (1 << row)])


n, m = map(int, input().split())
a = list(map(int, input().split()))
MOD = 998244353

sp = SparseTable(a, n)
to_ind = {}
for i in range(n):
    to_ind[a[i]] = i

dp = [[-1]*(n+1) for i in range(n+1)]


def solve(l, r):
    if dp[l][r] != -1:
        return dp[l][r]
    if l == r:
        dp[l][r] = 1
        return 1
    if r - l == 1:
        dp[l][r] = 1
        return 1
    ind = to_ind[sp.query(l, r)]
    res1 = 0
    res2 = 0
    for i in range(ind+1, r+1):
        res1 += solve(ind+1, i) * solve(i, r)
        res1 %= MOD
    for i in range(l, ind+1):
        res2 += solve(l, i) * solve(i, ind)
        res2 %= MOD
    dp[l][r] = max(res1, 1)*max(res2, 1)
    dp[l][r] %= MOD
    return dp[l][r]

print(solve(0, n) % MOD)
  	   		     	 		 			  	 		