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;
const int MOD = 998244353;
template <typename T>
static ostream& operator<<(ostream& stream, const vector<T>& t);
template <typename T>
static istream& read(T, T, istream& = cin);
template <typename T>
static istream& operator>>(istream& stream, vector<T>& t) {
  return read((t).begin(), (t).end(), stream);
}
template <typename T>
static istream& read(T b, T e, istream& stream) {
  for (T it = b; it != e; ++it) stream >> *it;
  return stream;
}
template <typename T>
static void print(T x, string end = "\n") {
  cout << x << end;
}
const int N = 505;
long long dp[N][N];
void _() {
  int n, m;
  cin >> n >> m;
  vector<int> arr(n);
  cin >> arr;
  for (int i = 0; i < n; ++i) {
    dp[i][i] = 1;
  }
  for (int s = 2; s <= n; ++s)
    for (int l = 0; l <= n - s; ++l) {
      int r = l + s - 1;
      int smallest = l;
      for (int i = l; i <= r; ++i)
        if (arr[i] < arr[smallest]) smallest = i;
      long long left_stuff = 1;
      if (l < smallest) left_stuff *= dp[l][smallest - 1];
      long long right_stuff = 1;
      if (r > smallest) right_stuff *= dp[smallest + 1][r];
      for (int i = smallest - 1; i >= l; --i) {
        if (i == l)
          left_stuff += dp[l][smallest - 1];
        else
          left_stuff += dp[i][smallest - 1] * dp[l][i - 1] % MOD;
      }
      left_stuff %= MOD;
      for (int i = smallest + 1; i <= r; ++i) {
        if (i == r)
          right_stuff += dp[smallest + 1][r];
        else
          right_stuff += dp[smallest + 1][i] * dp[i + 1][r] % MOD;
      }
      right_stuff %= MOD;
      dp[l][r] = right_stuff * left_stuff % MOD;
    }
  print(dp[0][n - 1]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed << setprecision(15);
  _();
}