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 long long md = 998244353;
long long n, m, a[1000005], dp[505][505];
int left_pt[505], right_pt[505];
int rev_l[1000005], rev_r[1000005];
vector<int> left_idx, right_idx, occurrences[505];
long long solve(int x, int y) {
  if (x > y) return 1;
  if (~dp[rev_l[x]][rev_r[y]]) return dp[rev_l[x]][rev_r[y]];
  int mn = 1e9;
  for (int i = x; i <= y; ++i)
    if (mn > a[i]) mn = a[i];
  if (left_pt[mn] < x || right_pt[mn] > y) return dp[rev_l[x]][rev_r[y]] = 0;
  long long ans = 1;
  long long left_sum = 0;
  for (int _l = lower_bound(left_idx.begin(), left_idx.end(), left_pt[mn]) -
                left_idx.begin();
       _l >= 0 && left_idx[_l] >= x; --_l) {
    int l = left_idx[_l];
    (left_sum += solve(x, l - 1) * solve(l, left_pt[mn] - 1)) %= md;
  }
  for (int i = 1; i < occurrences[mn].size(); ++i)
    (ans *= solve(occurrences[mn][i - 1] + 1, occurrences[mn][i] - 1)) %= md;
  long long right_sum = 0;
  for (int _r = lower_bound(right_idx.begin(), right_idx.end(), right_pt[mn]) -
                right_idx.begin();
       _r < right_idx.size() && right_idx[_r] <= y; ++_r) {
    int r = right_idx[_r];
    (right_sum += solve(r + 1, y) * solve(right_pt[mn] + 1, r)) %= md;
  }
  (ans *= left_sum * right_sum % md) %= md;
  return dp[rev_l[x]][rev_r[y]] = ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> m >> n;
  vector<int> arr;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    if (a[i] != a[i - 1]) arr.push_back(a[i]);
  }
  n = 0;
  for (auto i : arr) a[++n] = i;
  for (int i = 1; i <= n; ++i) occurrences[a[i]].push_back(i);
  for (int i = 1; i <= n; ++i) right_pt[a[i]] = i;
  for (int i = n; i >= 1; --i) left_pt[a[i]] = i;
  for (int i = 1; i <= m; ++i) {
    sort(occurrences[i].begin(), occurrences[i].end());
    left_idx.push_back(left_pt[i]);
    right_idx.push_back(right_pt[i]);
    rev_l[left_pt[i]] = i;
    rev_r[right_pt[i]] = i;
  }
  sort(left_idx.begin(), left_idx.end());
  sort(right_idx.begin(), right_idx.end());
  memset(dp, -1, sizeof(dp));
  dp[0][0] = dp[0][1] = dp[1][0] = 0;
  cout << solve(1, n);
}