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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int m, n;
int a[510];
int b[510][510];
int f[510][510];
int ff(int l, int r) {
  if (f[l][r] != -1) return f[l][r];
  if (l >= r - 1) return f[l][r] = 1;
  int S = 0;
  int T = 0;
  int p = b[l][r];
  for (int k = l; k <= p; k++)
    (S += (long long)ff(l, k) * ff(k, p) % 998244353) %= 998244353;
  for (int k = p + 1; k <= r; k++)
    (T += (long long)ff(p + 1, k) * ff(k, r) % 998244353) %= 998244353;
  return f[l][r] = (long long)S * T % 998244353;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> m >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
  }
  for (int i = 0; i < n; i++) {
    int p = -1;
    for (int j = i; j < n; j++) {
      if (p == -1 || a[j] < a[p]) p = j;
      b[i][j + 1] = p;
    }
  }
  memset(f, -1, sizeof f);
  cout << ff(0, n) << endl;
  return 0;
}