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 warning(disable : 4996)
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("-ffloat-store")
using namespace std;
clock_t time_p = clock();
void aryanc403() {
  time_p = clock() - time_p;
  cerr << "Time Taken : " << (float)(time_p) / CLOCKS_PER_SEC << "\n";
}
const long long int INF = 0xFFFFFFFFFFFFFFFL;
long long int seed;
mt19937 rng(seed = chrono::steady_clock::now().time_since_epoch().count());
inline long long int rnd(long long int l = 0, long long int r = INF) {
  return uniform_int_distribution<long long int>(l, r)(rng);
}
class CMP {
 public:
  bool operator()(pair<long long int, long long int> a,
                  pair<long long int, long long int> b) {
    return !(a.first < b.first || a.first == b.first && a.second <= b.second);
  }
};
void add(map<long long int, long long int> &m, long long int x,
         long long int cnt = 1) {
  auto jt = m.find(x);
  if (jt == m.end())
    m.insert({x, cnt});
  else
    jt->second += cnt;
}
void del(map<long long int, long long int> &m, long long int x,
         long long int cnt = 1) {
  auto jt = m.find(x);
  if (jt->second <= cnt)
    m.erase(jt);
  else
    jt->second -= cnt;
}
bool cmp(const pair<long long int, long long int> &a,
         const pair<long long int, long long int> &b) {
  return a.first < b.first || (a.first == b.first && a.second < b.second);
}
const long long int mod = 998244353LL;
const long long int maxN = 500;
long long int T, n, i, j, k, in, cnt, l, r, u, v, x, y;
long long int m;
string s;
vector<long long int> a, b;
long long int dp[2 * maxN + 5][2 * maxN + 5];
long long int frst[maxN + 5], lst[maxN + 5];
long long int solve(long long int l, long long int r) {
  if (l > r || r <= 0 || l > m) return 1;
  if (l == r) return frst[a[l]] == lst[a[l]];
  long long int &ans = dp[l][r];
  if (ans != -1) return ans;
  long long int i, j;
  long long int mn = INF;
  for (i = (l); i <= (r); ++i) mn = min(mn, a[i]);
  if (frst[mn] < l || r < lst[mn]) return ans = 0;
  long long int cnt = 0;
  i = l;
  while (a[i] != mn) i++;
  for (j = (l); j <= (i); ++j) cnt += (solve(l, j - 1) * solve(j, i - 1)) % mod;
  ans = cnt % mod;
  long long int pvr = i;
  i++;
  while (i <= r) {
    if (a[i] == mn) {
      ans = (ans * solve(pvr + 1, i - 1)) % mod;
      pvr = i;
    }
    i++;
  }
  cnt = 0;
  for (j = (pvr); j <= (r); ++j)
    cnt += (solve(pvr + 1, j) * solve(j + 1, r)) % mod;
  cnt %= mod;
  ans = (ans * cnt) % mod;
  ;
  return ans;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  {
    cin >> n >> m;
    a.clear();
    a.reserve(m);
    a.push_back(0);
    for (i = 0; i < (m); ++i) {
      cin >> in;
      a.push_back(in);
    }
    a.resize(unique(a.begin(), a.end()) - a.begin());
    for (i = (1); i <= (n); ++i) frst[i] = INF;
    m = (long long int)a.size();
    m--;
    cnt = 0;
    for (i = (1); i <= (m); ++i) {
      in = a[i];
      frst[in] = min(frst[in], i);
      lst[in] = max(lst[in], i);
    }
    if (m > 2 * n) {
      cout << 0 << "\n";
      return 0;
    };
    memset(dp, -1, (2 * maxN + 5) * (2 * maxN + 5) * sizeof(long long int));
    cout << solve(1, m) << "\n";
  }
  aryanc403();
  return 0;
}