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;
string to_string(const string& s) { return '"' + s + '"'; }
string to_string(const bool b) { return (b ? "true" : "false"); }
string to_string(const char* s) { return to_string((string)s); }
string to_string(const char c) {
  string s = "'";
  s += c;
  s += "'";
  return s;
}
template <class T, class U>
string to_string(const pair<T, U>& p) {
  string s = "(";
  s += to_string(p.first);
  s += ", ";
  s += to_string(p.second);
  s += ")";
  return s;
}
template <class T>
string to_string(const T& a) {
  bool first = true;
  string s = "{";
  for (const auto& x : a) {
    if (!first) {
      s += ", ";
    }
    first = false;
    s += to_string(x);
  }
  s += "}";
  return s;
}
void debug_out() { cerr << '\n'; }
template <class Head, class... Tail>
void debug_out(Head head, Tail... tail) {
  cerr << ' ' << to_string(head);
  debug_out(tail...);
}
using ll = long long;
using ld = long double;
const int mod = 998244353;
const int inf = 2e9;
const ll inf64 = 7e18l;
int n, m;
vector<int> a;
vector<vector<int>> memo;
int mult(int x, int y) { return 1ll * x * y % mod; }
int add(int x, int y) {
  x += y;
  if (x >= mod) {
    return x - mod;
  }
  return x;
}
void add_self(int& x, int y) { x = add(x, y); }
int solve(int l, int r) {
  if (l >= r) {
    return 1;
  }
  assert(l >= 0 && r >= 0);
  if (memo[l][r] != -1) {
    return memo[l][r];
  }
  int ind = min_element(a.begin() + l, a.begin() + r + 1) - a.begin();
  int left = 0, right = 0;
  for (int l2 = l - 1; l2 < ind; ++l2) {
    add_self(left, mult(solve(l, l2), solve(l2 + 1, ind - 1)));
  }
  for (int r2 = ind + 1; r2 <= r + 1; ++r2) {
    add_self(right, mult(solve(ind + 1, r2 - 1), solve(r2, r)));
  }
  memo[l][r] = mult(left, right);
  return memo[l][r];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(20);
  cin >> n >> m;
  a.resize(n);
  memo.resize(n, vector<int>(n, -1));
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  cout << solve(0, n - 1) << '\n';
  return 0;
}