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 mod = 998244353;
long long dp[1010][1010];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<int> a({0});
  int minleft[n + 1], minright[n + 1];
  vector<int> pos[n + 1];
  fill(minleft, minleft + n + 1, -1);
  fill(minright, minright + n + 1, -1);
  for (int i = 1; i <= m; i++) {
    int k;
    cin >> k;
    if (a.back() != k) a.push_back(k);
  }
  if (a.size() > 2 * n + 1) {
    cout << 0 << endl;
    return 0;
  }
  m = a.size() - 1;
  for (int i = 1; i <= m; i++) {
    if (minleft[a[i]] == -1) minleft[a[i]] = i;
    minright[a[i]] = i;
    pos[a[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    dp[minleft[i]][minright[i]] = 1;
  }
  for (int i = 1; i <= m + 1; i++) {
    dp[i][i - 1] = 1;
  }
  for (int d = 2; d <= m; d++) {
    for (int l = 1, r = l + d - 1; r <= m; l++, r++) {
      int mn = n + 1;
      for (int i = l; i <= r; i++) {
        if (a[i] < mn) mn = a[i];
      }
      if (minleft[mn] < l or minright[mn] > r) continue;
      long long lans = 0, rans = 0;
      for (int i = l; i <= minleft[mn]; i++) {
        lans += dp[l][i - 1] * dp[i][minleft[mn] - 1];
        lans %= mod;
      }
      for (int i = minright[mn]; i <= r; i++) {
        rans += dp[minright[mn] + 1][i] * dp[i + 1][r];
        rans %= mod;
      }
      dp[l][r] = lans * rans % mod;
      int cur = minleft[mn];
      if (minleft[mn] != minright[mn]) {
        for (auto x : pos[mn]) {
          (dp[l][r] *= dp[cur][x - 1]) %= mod;
          cur = x + 1;
        }
      }
    }
  }
  cout << dp[1][m] << '\n';
  return 0;
}