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 OO = 1e9;
const double EPS = 1e-9;
template <class T>
void _db(const char* dbStr, T e) {
  cout << dbStr << " = " << e << endl;
}
template <class T, class... L>
void _db(const char* dbStr, T e, L... r) {
  while (*dbStr != ',') cout << *dbStr++;
  cout << " = " << e << ',';
  _db(dbStr + 1, r...);
}
template <class S, class T>
ostream& operator<<(ostream& o, const map<S, T>& v) {
  o << "[";
  int i = 0;
  for (const pair<S, T>& pr : v)
    o << (!i++ ? "" : ", ") << "{" << pr.first << " : " << pr.second << "}";
  return o << "]";
}
template <template <class, class...> class S, class T, class... L>
ostream& operator<<(ostream& o, const S<T, L...>& v) {
  o << "[";
  int i = 0;
  for (const auto& e : v) o << (!i++ ? "" : ", ") << e;
  return o << "]";
}
template <class S, class T>
ostream& operator<<(ostream& o, const pair<S, T>& pr) {
  return o << "(" << pr.first << ", " << pr.second << ")";
}
ostream& operator<<(ostream& o, const string& s) {
  for (const char& c : s) o << c;
  return o;
}
template <class T>
using V = vector<T>;
template <class T>
using VV = V<V<T>>;
template <class T>
using VVV = VV<V<T>>;
using ll = long long;
using pii = pair<int, int>;
using vi = V<int>;
using vii = V<pii>;
using vvi = VV<int>;
using mii = map<int, int>;
using umii = unordered_map<int, int>;
using si = set<int>;
using usi = unordered_set<int>;
const int MX = 505, MOD = 998244353;
int n, m, mem[MX * 2][MX * 2];
vi c, cLong, hi, lo;
int dp(int st, int en) {
  if (st > en) return 1;
  int& ret = mem[st][en];
  if (ret != -1) return ret;
  int mn = n + 1;
  for (int i = st; i <= en; ++i)
    if (c[i] < mn) mn = c[i];
  ret = 0;
  if (lo[mn] < st || hi[mn] > en) return ret;
  ll mult = 1;
  for (int k = lo[mn]; k <= hi[mn]; ++k)
    if (c[k] != mn)
      for (int l = k; l <= hi[mn]; ++l)
        if (l + 1 > hi[mn] || c[l + 1] == mn) {
          mult = (mult * dp(k, l)) % MOD;
          k = l;
          break;
        }
  ll sumL = 0;
  for (int i = st; i <= lo[mn]; ++i)
    sumL = (sumL + (dp(st, i - 1) * 1LL * dp(i, lo[mn] - 1)) % MOD) % MOD;
  for (int j = hi[mn]; j <= en; ++j)
    ret = (ret + sumL * (dp(hi[mn] + 1, j) * 1LL * dp(j + 1, en) % MOD)) % MOD;
  ret = (ret * mult) % MOD;
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cout.precision(1);
  cin.tie(0);
  cin >> n >> m;
  cLong.assign(m + 1, 0);
  for (int i = 1; i <= m; ++i) cin >> cLong[i];
  true;
  c.push_back(0);
  for (int i = 1; i <= m; ++i)
    if (cLong[i] != cLong[i - 1]) c.push_back(cLong[i]);
  hi.assign(n + 1, 0), lo.assign(n + 1, 0);
  for (int i = 1; i < int(c.size()); ++i) {
    hi[c[i]] = i;
    if (!lo[c[i]]) lo[c[i]] = i;
  }
  if (int(c.size()) - 1 > 2 * n) {
    cout << "0\n";
    return 0;
  }
  true;
  memset(mem, -1, sizeof(mem));
  cout << dp(1, int(c.size()) - 1) << '\n';
  return 0;
}