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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
void debug_out() { cerr << endl; }
template <class T>
ostream& prnt(ostream& out, T v) {
  out << v.size() << '\n';
  for (auto e : v) out << e << ' ';
  return out;
}
template <class T>
ostream& operator<<(ostream& out, vector<T> v) {
  return prnt(out, v);
}
template <class T>
ostream& operator<<(ostream& out, set<T> v) {
  return prnt(out, v);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, map<T1, T2> v) {
  return prnt(out, v);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> p) {
  return out << '(' << p.first << ' ' << p.second << ')';
}
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << H;
  debug_out(T...);
}
const long long N = 1000005;
const long long MOD = 998244353;
const long long OFF = 1e7 + 87;
long long n, a[N], x[N], m, dp[5000][5000], last;
vector<long long> pos[N];
vector<pair<long long, long long> > v;
inline int tr(int val) {
  if (x[val] == 0) {
    x[val] = ++last;
  }
  return x[val];
}
long long solve(long long l, long long r, int save = 1) {
  if (l > r) return 1;
  if (!save || dp[tr(l)][tr(r)] == -1) {
    if (l == r) {
      long long pmin = *pos[a[l]].begin(), pmax = pos[a[l]].back();
      if (save) dp[tr(l)][tr(r)] = 0;
      if (pmin < l || r < pmax) return 0;
      if (save) dp[tr(l)][tr(r)] = 1;
      return 1;
    } else {
      long long mn = l;
      for (long long i = l; i <= r; i++) {
        if (a[i] < a[mn]) mn = i;
      }
      long long pmin = *pos[a[mn]].begin(), pmax = pos[a[mn]].back();
      if (pmin < l || r < pmax) return 0;
      long long sum_mid = 1;
      for (long long i = 1; i < pos[a[mn]].size(); i++)
        sum_mid =
            (sum_mid * solve(pos[a[mn]][i - 1] + 1, pos[a[mn]][i] - 1, 0)) %
            MOD;
      long long sum_left = 0;
      for (long long i = l; i <= pmin; i++) {
        sum_left += ((solve(l, i - 1) * solve(i, pmin - 1)) % MOD);
        sum_left %= MOD;
      }
      long long ret = 0;
      for (long long j = pmax; j <= r; j++)
        ret += (((sum_mid * sum_left) % MOD) *
                ((solve(pmax + 1, j) * solve(j + 1, r)) % MOD)) %
               MOD;
      ret %= MOD;
      if (save) dp[tr(l)][tr(r)] = ret;
      return ret;
    }
  }
  return dp[tr(l)][tr(r)];
}
bool check() { return 1; }
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  for (long long i = 1; i <= m; i++) cin >> a[i];
  for (int i = 1, j = 1; i <= m; i++) {
    a[i] = a[j];
    while (j <= m && a[i] == a[j]) j++;
    if (j > m) {
      m = i;
      break;
    }
  }
  for (long long i = 1; i <= m; i++) pos[a[i]].push_back(i);
  if (!check()) return cout << "0\n", 0;
  memset(dp, -1, sizeof dp);
  cout << solve(1, m) << '\n';
}