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;
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  bool first = true;
  for (const auto& j : v) {
    if (!first) {
      os << " ";
    }
    first = false;
    os << j;
  }
  os << "\n";
  return os;
}
template <typename T>
istream& operator>>(istream& is, vector<T>& v) {
  for (auto& elem : v) {
    is >> elem;
  }
  return is;
}
bool is_p(int n) {
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
void solve() {
  int n, m;
  cin >> n >> m;
  vector<int> a(n);
  cin >> a;
  vector<vector<long long> > dp(n + 1, vector<long long>(n + 1, 1));
  for (int l = 2; l <= n; l++) {
    for (int j = 0; j <= n - l; j++) {
      int mi = j;
      for (int i = j; i < j + l; i++) {
        if (a[i] < a[mi]) {
          mi = i;
        }
      }
      long long x = 0;
      for (int i = j; i <= mi; i++) {
        x = (x + (dp[j][i - j] * dp[i][mi - i]) % 998244353) % 998244353;
      }
      long long y = 0;
      for (int i = mi; i < j + l; i++) {
        y = (y + (dp[mi + 1][i - mi] * dp[i + 1][j + l - 1 - i]) % 998244353) %
            998244353;
      }
      dp[j][l] = (x * y) % 998244353;
    }
  }
  cout << dp[0][n];
}
int main() {
  ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  if (0) {
    int tests;
    cin >> tests;
    while (tests--) {
      solve();
      cout << "\n";
    }
  } else {
    solve();
    cout << endl;
  }
  return 0;
}