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;
long long mod = 998244353;
template <class C>
void mini(C& a, C b) {
  a = min(a, b);
}
template <class C>
void maxi(C& a, C b) {
  a = max(a, b);
}
void a() {
  string s;
  cin >> s;
  reverse(s.begin(), s.end());
  int n = s.size();
  int cerosMenosUnos = 0;
  string res;
  for (int i = 0; i < n; i++) {
    if (cerosMenosUnos == 0 or s[i] == '0')
      res.push_back('0');
    else
      res.push_back('1');
    if (s[i] == '0')
      cerosMenosUnos++;
    else
      cerosMenosUnos--;
    maxi(cerosMenosUnos, 0);
  }
  reverse(res.begin(), res.end());
  cout << res << '\n';
}
struct segmentMin {
  segmentMin(vector<pair<int, int> >& values);
  void asignar(int i, pair<int, int> x);
  pair<int, int> Minr(int a, int b);
  int n;
  vector<pair<int, int> > tree;
};
segmentMin::segmentMin(vector<pair<int, int> >& values) {
  n = values.size();
  tree = vector<pair<int, int> >(2 * n);
  for (int i = 0; i < n; i++) tree[n + i] = values[i];
  for (int(i) = n - 1; i >= 1; i--) tree[i] = min(tree[2 * i], tree[2 * i + 1]);
}
void segmentMin::asignar(int i, pair<int, int> x) {
  i += n;
  tree[i] = x;
  for (i /= 2; i >= 1; i /= 2) tree[i] = min(tree[2 * i], tree[2 * i + 1]);
}
pair<int, int> segmentMin::Minr(int a, int b) {
  a += n;
  b += n;
  pair<int, int> Minimo = make_pair(100000000, 100000000);
  while (a <= b) {
    if (a % 2 == 1) mini(Minimo, tree[a++]);
    if (b % 2 == 0) mini(Minimo, tree[b--]);
    a /= 2;
    b /= 2;
  }
  return Minimo;
}
void b() {
  int n;
  cin >> n >> n;
  vector<int> colores(n);
  for (int i = 0; i < n; i++) cin >> colores[i];
  vector<pair<int, int> > values;
  for (int i = 0; i < n; i++) values.push_back(make_pair(colores[i], i));
  segmentMin siguienteColor(values);
  vector<vector<long long> > dp(n + 1, vector<long long>(n + 1));
  for (int i = 0; i < n; i++) dp[i][i] = 1;
  for (int(i) = 1; i <= n; i++) {
    dp[i][i - 1] = 1;
  }
  for (int(len) = 1; len <= n - 1; len++) {
    for (int i = 0; i <= n - len - 1; i++) {
      int j = i + len;
      int m = siguienteColor.Minr(i, j).second;
      long long aux = 0;
      if (m > i) {
        for (int k = i; k <= m; k++) {
          aux += dp[i][max(0, k - 1)] * dp[k][m - 1];
          aux %= mod;
        }
      } else
        aux = 1;
      if (m < n - 1) {
        for (int k = m; k <= j; k++) {
          dp[i][j] += dp[m + 1][k] * dp[k + 1][j];
          dp[i][j] %= mod;
        }
      } else
        dp[i][j] = 1;
      dp[i][j] *= aux;
      dp[i][j] %= mod;
    }
  }
  cout << dp[0][n - 1] << '\n';
}
int main() {
  b();
  return 0;
}