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;
template <class T, class U>
void ckmin(T &a, U b) {
  if (a > b) a = b;
}
template <class T, class U>
void ckmax(T &a, U b) {
  if (a < b) a = b;
}
int N, M;
int arr[1000013];
pair<int, int> range[513];
int parent[513];
vector<int> edge[513];
bitset<513> vis[513];
int stor[20][1000013];
int dp[513][513];
vector<int> ord;
int ans = 1;
int ad(int a, int b) {
  a += b;
  if (a >= 998244353) a -= 998244353;
  return a;
}
int sb(int a, int b) {
  a -= b;
  if (a < 0) a += 998244353;
  return a;
}
int mul(int a, int b) { return (long long)a * b % 998244353; }
bool cmp(int a, int b) { return range[a].first < range[b].first; }
int solve(int lt, int rt) {
  if (lt > rt) return 1;
  if (vis[lt][rt]) return dp[lt][rt];
  vis[lt][rt] = true;
  int mn = lt;
  for (auto i = (lt); i < (rt + 1); i++) {
    if (ord[i] < ord[mn]) mn = i;
  }
  int sl = 0, sr = 0;
  for (auto i = (lt); i < (mn + 1); i++) {
    sl = ad(sl, mul(solve(lt, i - 1), solve(i, mn - 1)));
  }
  for (auto i = (mn); i < (rt + 1); i++) {
    sr = ad(sr, mul(solve(mn + 1, i), solve(i + 1, rt)));
  }
  dp[lt][rt] = mul(sl, sr);
  return dp[lt][rt];
}
void dfs(int u) {
  sort((edge[u]).begin(), (edge[u]).end(), cmp);
  for (int v : edge[u]) {
    dfs(v);
  }
  for (auto i = (0); i < (((int)(edge[u]).size())); i++) {
    ord.push_back(edge[u][i]);
    if (i == ((int)(edge[u]).size()) - 1 ||
        range[edge[u][i + 1]].first != range[edge[u][i]].second + 1) {
      int n = ((int)(ord).size());
      for (auto i = (0); i < (n); i++) {
        for (auto j = (i); j < (n); j++) {
          vis[i][j] = false;
        }
      }
      ans = mul(ans, solve(0, n - 1));
      ord.clear();
    }
  }
  return;
}
int query(int l, int r) {
  int sz = 31 - __builtin_clz(r - l + 1);
  return min(stor[sz][l], stor[sz][r - (1 << sz) + 1]);
}
int32_t main() {
  cout << fixed << setprecision(12);
  cerr << fixed << setprecision(4);
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> M;
  N++;
  for (auto i = (1); i < (N + 1); i++) {
    range[i] = {M, -1};
  }
  range[0] = {0, M + 1};
  for (auto i = (1); i < (M + 1); i++) {
    int x;
    cin >> x;
    arr[i] = x;
    ckmax(range[x].second, i);
    ckmin(range[x].first, i);
  }
  M += 2;
  for (auto i = (0); i < (M); i++) {
    stor[0][i] = arr[i];
  }
  for (auto i = (1); i < (20); i++) {
    for (auto j = (0); j < (M - (1 << i) + 1); j++) {
      stor[i][j] = min(stor[i - 1][j], stor[i - 1][j + (1 << (i - 1))]);
    }
  }
  for (auto i = (0); i < (N); i++) {
    int mn = query(range[i].first, range[i].second);
    if (mn < i) {
      cout << "0\n";
      return 0;
    }
  }
  for (auto i = (N)-1; i >= (1); i--) {
    for (auto j = (0); j < (i); j++) {
      if (range[j].second < range[i].first || range[j].first > range[i].second)
        continue;
      if (range[j].first < range[i].first &&
          range[i].second < range[j].second) {
        parent[i] = j;
      }
    }
    edge[parent[i]].push_back(i);
  }
  dfs(0);
  cout << ans << '\n';
  return 0;
}