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 inf = 1e9 + 7;
const long long INF = 1e18L + 7;
template <class n, class second>
ostream &operator<<(ostream &p, pair<n, second> x) {
  return p << "<" << x.first << ", " << x.second << ">";
}
template <class n>
auto operator<<(ostream &p, n y) ->
    typename enable_if<!is_same<n, string>::value,
                       decltype(y.begin(), p)>::type {
  int o = 0;
  p << "{";
  for (auto c : y) {
    if (o++) p << ", ";
    p << c;
  }
  return p << "}";
}
void dor() { cerr << '\n'; }
template <class n, class... second>
void dor(n p, second... y) {
  cerr << p << " ";
  dor(y...);
}
template <class n, class second>
void mini(n &p, second y) {
  if (p > y) p = y;
}
template <class n, class second>
void maxi(n &p, second y) {
  if (p < y) p = y;
}
const int mod = 998244353;
const int N = 507;
int n, m;
int tab[N];
int dp[N][N][3];
int vis[N][N][3];
int pie[N];
int dru[N];
void solve(int a, int b, int t) {
  if (vis[a][b][t]) return;
  vis[a][b][t] = 1;
  if (a > b) {
    if (t == 2)
      dp[a][b][t] = 1;
    else
      dp[a][b][t] = 0;
    return;
  }
  int col = inf;
  int ind = -1;
  if (t == 2) {
    for (int i = a; i <= b; ++i) {
      if (col > tab[i]) {
        col = tab[i];
        ind = i;
      }
    }
    solve(a, ind, 1);
    solve(ind, b, 0);
    dp[a][b][t] = (long long)dp[a][ind][1] * dp[ind][b][0] % mod;
  } else if (t == 0) {
    for (int i = a; i <= b; ++i) {
      solve(a + 1, i, 2);
      solve(i + 1, b, 2);
      dp[a][b][t] += (long long)dp[a + 1][i][2] * dp[i + 1][b][2] % mod;
      if (dp[a][b][t] >= mod) dp[a][b][t] -= mod;
    }
  } else if (t == 1) {
    for (int i = a; i <= b; ++i) {
      solve(a, i - 1, 2);
      solve(i, b - 1, 2);
      dp[a][b][t] += (long long)dp[a][i - 1][2] * dp[i][b - 1][2] % mod;
      if (dp[a][b][t] >= mod) dp[a][b][t] -= mod;
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) {
    cin >> tab[i];
  }
  solve(1, n, 2);
  cout << dp[1][n][2] << '\n';
}