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>
template <class C>
inline void log_c(const C& c) {}
template <class C>
inline int sz(const C& c) {
  return static_cast<int>(c.size());
}
using namespace std;
using pii = pair<int, int>;
using num = int64_t;
using pll = pair<num, num>;
const std::string eol = "\n";
void logdp(const vector<num>& dp, int n) {
  for (int k = 0; k < (n + 1); ++k) {
    for (int j = 0; j < (n + 1); ++j) clog << setw(4) << dp[k * (n + 1) + j];
    clog << endl;
  }
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  const num mod = 998244353;
  int n, m;
  cin >> n >> m;
  vector<int> pos(n + 1);
  vector<int> c(n);
  for (int k = 0; k < (m); ++k) {
    cin >> c[k];
    pos[c[k]] = k;
  }
  vector<num> dp((n + 1) * (n + 1));
  for (int k = 0; k < (n + 1); ++k) dp[k * (n + 1) + k] = 1;
  for (int col = m; col; --col) {
    int p = pos[col];
    c[p] = 0;
    for (int l = p; l >= 0 && !c[l]; --l) {
      num ax = 0;
      for (int x = p; x >= l; --x) {
        num lmul = dp[l * (n + 1) + x] * dp[x * (n + 1) + p] % mod;
        ax = (ax + lmul) % mod;
      }
      dp[l * (n + 1) + p + 1] = ax;
    }
    for (int r = p + 1; r <= n && !c[r - 1]; ++r) {
      num ax = 0;
      for (int y = p + 1; y <= r; ++y) {
        num rmul = dp[(p + 1) * (n + 1) + y] * dp[y * (n + 1) + r] % mod;
        ax = (ax + rmul) % mod;
      }
      dp[p * (n + 1) + r] = ax;
    }
    for (int l = p - 1; l >= 0 && !c[l]; --l) {
      for (int r = p + 2; r <= n && !c[r - 1]; ++r) {
        dp[l * (n + 1) + r] =
            dp[l * (n + 1) + p + 1] * dp[p * (n + 1) + r] % mod;
      }
    }
  }
  cout << dp[n] << endl;
  return 0;
}