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;
long long n, m;
vector<long long> values;
long long dp[1010][1010];
vector<long long> amount[510];
int main() {
  cin >> n >> m;
  values.push_back(1e12);
  for (long long i = 1; i <= m; i++) {
    long long x;
    cin >> x;
    if (values.back() != x) {
      values.push_back(x);
    }
  }
  m = values.size() - 1;
  if (m > 2 * n) {
    cout << "0\n";
    return 0;
  }
  for (long long i = 1; i <= m; i++) {
    amount[values[i]].push_back(i);
  }
  for (long long i = 0; i < 1010; i++) {
    for (long long j = 0; j < 1010; j++) {
      dp[i][j] = 1;
    }
  }
  for (long long l = 1; l <= m; l++) {
    for (long long start = 1; start + l - 1 <= m; start++) {
      long long minimum_colour = 1e12;
      for (long long i = start; i < start + l; i++) {
        minimum_colour = min(minimum_colour, values[i]);
      }
      long long minimum_index = 1e12, maximum_index = -1e12;
      for (auto i : amount[minimum_colour]) {
        minimum_index = min(minimum_index, i);
        maximum_index = max(maximum_index, i);
      }
      if (minimum_index < start || maximum_index > start + l - 1) {
        dp[start][l] = 0;
        continue;
      }
      long long left = 0, right = 0;
      for (long long u = start; u <= minimum_index; u++) {
        left += dp[start][u - start] * dp[u][minimum_index - u];
        left %= mod;
      }
      for (long long u = maximum_index + 1; u <= start + l; u++) {
        right +=
            dp[maximum_index + 1][u - maximum_index - 1] * dp[u][start + l - u];
        right %= mod;
      }
      dp[start][l] = right * left;
      dp[start][l] %= mod;
      for (long long i = 0; i < amount[minimum_colour].size() - 1; i++) {
        dp[start][l] *=
            dp[amount[minimum_colour][i] + 1]
              [amount[minimum_colour][i + 1] - amount[minimum_colour][i] - 1];
        dp[start][l] %= mod;
      }
    }
  }
  cout << dp[1][m] << '\n';
  return 0;
}