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 mod = 998244353;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<int> c(m);
  vector<int> w;
  for (int i = 0; i < m; i++) {
    cin >> c[i];
    c[i]--;
    if (!i || c[i] != c[i - 1]) w.push_back(i);
  }
  if (w.size() > n * 2) {
    cout << "0\n";
    return 0;
  }
  m = w.size();
  vector<vector<int> > o(n);
  for (int i = 0; i < m; i++) {
    o[c[w[i]]].push_back(i);
  }
  vector<vector<long long> > dp(m + 1, vector<long long>(m + 1, 1));
  for (int len = 1; len <= m; len++) {
    for (int l = 0; l + len <= m; l++) {
      int r = l + len;
      int mini = 1e9;
      for (int i = l; i < r; i++) mini = min(mini, c[w[i]]);
      int o1 = o[mini][0], o2 = o[mini].back();
      if (o1 < l || o2 >= r) {
        dp[l][r] = 0;
        continue;
      }
      long long left = 0, right = 0;
      for (int i = l; i <= o1; i++) {
        left += (dp[l][i] * dp[i][o1]) % mod;
        if (left >= mod) left -= mod;
      }
      for (int i = o2 + 1; i <= r; i++) {
        right += (dp[o2 + 1][i] * dp[i][r]) % mod;
        if (right >= mod) right -= mod;
      }
      dp[l][r] = (left * right) % mod;
      for (int i = 0; i < o[mini].size() - 1; i++) {
        dp[l][r] = (dp[l][r] * dp[o[mini][i] + 1][o[mini][i + 1]]) % mod;
      }
    }
  }
  cout << dp[0][m] << "\n";
  return 0;
}