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;
using namespace std::chrono;
long long mod = 998244353;
long long powm(long long a, long long b, long long md = mod) {
  b = b / md + b % md;
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % md;
    a = (a * a) % md;
    b = b >> 1;
  }
  return ans;
}
long long st[501][10];
long long n;
long long get_min(long long l, long long r) {
  l--;
  r--;
  long long lt = log2(r - l + 1);
  return min(st[l][lt], st[r - (1 << lt) + 1][lt]);
}
long long dp[501][501];
long long in[501];
long long get(long long a, long long b) {
  if (b <= a) return 1;
  if (dp[a][b]) return dp[a][b];
  if (a == b) return 1;
  long long mini = get_min(a, b);
  mini = in[mini];
  long long left = get(a, mini - 1);
  for (long long i = mini - 1; i >= a; i--) {
    left = (left + (get(a, i - 1) * get(i, mini - 1)) % mod) % mod;
  }
  long long right = get(mini + 1, b);
  for (long long i = mini + 1; i <= b; i++) {
    right = (right + (get(mini + 1, i) * get(i + 1, b)) % mod) % mod;
  }
  dp[a][b] = (left * right) % mod;
  return dp[a][b];
}
void solve() {
  cin >> n >> n;
  for (long long i = 0; i < n; i++) {
    cin >> st[i][0];
    in[st[i][0]] = i + 1;
  }
  for (long long j = 1; j < 10; j++)
    for (long long i = 0; i + (1 << j) - 1 < n; i++) {
      st[i][j] = min(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
    }
  cout << get(1, n) << '\n';
}
signed main() {
  auto start = high_resolution_clock::now();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed << setprecision(15);
  long long t = 1;
  while (t--) solve();
  auto stop = high_resolution_clock::now();
  auto duration = duration_cast<microseconds>(stop - start);
  cerr << "duration.count()"
       << " is " << duration.count() << '\n';
  ;
  return 0;
}