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 int64_t MOD = 998244353;
template <typename _T>
inline void _DBG(const char* s, _T x);
template <typename _T, typename... args>
void _DBG(const char* s, _T x, args... a);
ostream& operator<<(ostream& out, const vector<int64_t>& a);
template <typename T>
ostream& operator<<(ostream& out, const vector<T>& a);
template <typename T>
istream& operator>>(istream& in, vector<T>& a);
int64_t calc_dp(vector<vector<int64_t>>& dp, vector<int64_t>& a, int64_t l,
                int64_t r) {
  if (r + 1 == l) {
    return 1;
  }
  assert(l <= r);
  if (dp[l][r]) {
    return dp[l][r];
  }
  if (l == r) {
    dp[l][r] = 1;
    return 1;
  }
  int64_t max_ind = min_element(a.begin() + l, a.begin() + r + 1) - a.begin();
  int64_t val1 = 0;
  for (int64_t i = l; i <= max_ind; ++i) {
    val1 = (val1 + calc_dp(dp, a, l, i - 1) * calc_dp(dp, a, i, max_ind - 1)) %
           MOD;
  }
  int64_t val2 = 0;
  for (int64_t j = max_ind; j <= r; ++j) {
    val2 = (val2 + calc_dp(dp, a, max_ind + 1, j) * calc_dp(dp, a, j + 1, r)) %
           MOD;
  }
  return dp[l][r] = (val1 * val2) % MOD;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int64_t n, m;
  cin >> n >> m;
  vector<int64_t> a(n);
  cin >> a;
  vector<vector<int64_t>> dp(n, vector<int64_t>(n));
  cout << calc_dp(dp, a, 0, n - 1) << "\n";
  return 0;
}
template <typename T>
istream& operator>>(istream& in, vector<T>& a) {
  for (auto it = a.begin(); it != a.end(); ++it) {
    in >> *it;
  }
  return in;
}
template <typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
  for (auto it = a.begin(); it != a.end(); ++it) {
    out << *it << "\n";
  }
  return out;
}
ostream& operator<<(ostream& out, const vector<int64_t>& a) {
  for (auto it = a.begin(); it != a.end(); ++it) {
    out << *it << " ";
  }
  return out;
}