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;
template <class T>
using min_heap = priority_queue<T, vector<T>, greater<T>>;
namespace input {
template <class T>
void re(complex<T>& x);
template <class T1, class T2>
void re(pair<T1, T2>& p);
template <class T>
void re(vector<T>& a);
template <class T, size_t SZ>
void re(array<T, SZ>& a);
template <class T>
void reA(T A[], int sz);
template <class T>
void re(T& x) {
  cin >> x;
}
void re(double& x) {
  string t;
  re(t);
  x = stod(t);
}
void re(long double& x) {
  string t;
  re(t);
  x = stold(t);
}
template <class Arg, class... Args>
void re(Arg& first, Args&... rest) {
  re(first);
  re(rest...);
}
template <class T1, class T2>
void re(pair<T1, T2>& p) {
  re(p.first, p.second);
}
template <class T>
void re(vector<T>& a) {
  for (int i = 0; i < ((int)a.size()); i++) re(a[i]);
}
template <class T, size_t SZ>
void re(array<T, SZ>& a) {
  for (int i = 0; i < (SZ); i++) re(a[i]);
}
template <class T>
void reA(T A[], int sz) {
  for (int i = 0; i < (sz); i++) re(A[i]);
}
void setupIO(const string& PROB = "") {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  if (PROB.length() != 0) {
    ifstream infile(PROB + ".in");
    if (infile.good()) {
      freopen((PROB + ".in").c_str(), "r", stdin);
      freopen((PROB + ".out").c_str(), "w", stdout);
    }
  }
}
}  // namespace input
using namespace input;
namespace output {
template <class T1, class T2>
void prD(const pair<T1, T2>& x);
template <class T, size_t SZ>
void prD(const array<T, SZ>& x);
template <class T>
void prD(const vector<T>& x);
template <class T>
void prD(const set<T>& x);
template <class T1, class T2>
void prD(const map<T1, T2>& x);
template <class T1, class T2>
void pr(const pair<T1, T2>& x);
template <class T, size_t SZ>
void pr(const array<T, SZ>& x);
template <class T>
void pr(const vector<T>& x);
template <class T>
void pr(const set<T>& x);
template <class T1, class T2>
void pr(const map<T1, T2>& x);
template <class T>
void prD(const T& x) {
  cout << x;
}
template <class Arg, class... Args>
void prD(const Arg& first, const Args&... rest) {
  prD(first);
  prD(rest...);
}
template <class T1, class T2>
void prD(const pair<T1, T2>& x) {
  prD("{", x.first, ", ", x.second, "}");
}
template <class T>
void prDContain(const T& x) {
  prD("{");
  bool fst = 1;
  for (const auto& a : x) prD(!fst ? ", " : "", a), fst = 0;
  prD("}");
}
template <class T, size_t SZ>
void prD(const array<T, SZ>& x) {
  prDContain(x);
}
template <class T>
void prD(const vector<T>& x) {
  prDContain(x);
}
template <class T>
void prD(const set<T>& x) {
  prDContain(x);
}
template <class T1, class T2>
void prD(const map<T1, T2>& x) {
  prDContain(x);
}
void psD() { prD("\n"); }
template <class Arg>
void psD(const Arg& first) {
  prD(first);
  psD();
}
template <class Arg, class... Args>
void psD(const Arg& first, const Args&... rest) {
  prD(first, " ");
  psD(rest...);
}
template <class T>
void pr(const T& x) {
  cout << x;
}
template <class Arg, class... Args>
void pr(const Arg& first, const Args&... rest) {
  pr(first);
  pr(rest...);
}
template <class T1, class T2>
void pr(const pair<T1, T2>& x) {
  pr(x.first, " ", x.second);
}
template <class T>
void prContain(const T& x) {
  bool fst = 1;
  for (const auto& a : x) pr(!fst ? " " : "", a), fst = 0;
}
template <class T, size_t SZ>
void pr(const array<T, SZ>& x) {
  prContain(x);
}
template <class T>
void pr(const vector<T>& x) {
  prContain(x);
}
template <class T>
void pr(const set<T>& x) {
  prContain(x);
}
template <class T1, class T2>
void pr(const map<T1, T2>& x) {
  prContain(x);
}
void ps() { pr("\n"); }
template <class Arg>
void ps(const Arg& first) {
  pr(first);
  ps();
}
template <class Arg, class... Args>
void ps(const Arg& first, const Args&... rest) {
  pr(first, " ");
  ps(rest...);
}
}  // namespace output
using namespace output;
int mod = 998244353;
long long dp[500][500];
long long transition[500];
int minVal[500];
int A[500];
long long getDP(int i, int j) {
  if (i >= j) return 1;
  return dp[i][j];
}
int main() {
  setupIO();
  int n, m;
  re(n, m);
  reA(A, m);
  for (int i = 0; i < (n); i++)
    for (int j = 0; j < (n); j++) dp[i][j] = 1;
  ;
  for (int i = 0; i < n; i++) transition[i] = 1;
  ;
  for (int length = 0; length < (n); length++) {
    for (int start = 0; start < (n - length); start++) {
      int end = start + length;
      if (length == 0) {
        dp[start][start] = 1;
        minVal[start] = start;
        continue;
      }
      int minIdx = minVal[start];
      if (A[end] < A[minIdx]) {
        minVal[start] = end;
        minIdx = end;
        long long newT = 0;
        for (int i = start; i < (minIdx + 1); i++) {
          newT = (newT + getDP(start, i - 1) * getDP(i, minIdx - 1)) % mod;
        }
        transition[start] = newT;
        dp[start][end] = transition[start];
      } else {
        dp[start][end] = 0;
        for (int i = minIdx; i < (end + 1); i++) {
          dp[start][end] =
              (dp[start][end] + transition[start] * getDP(minIdx + 1, i) % mod *
                                    getDP(i + 1, end)) %
              mod;
        }
      }
    }
  }
  ps(dp[0][n - 1]);
  return 0;
}