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 <typename S, typename T>
ostream& operator<<(ostream& out, pair<S, T> const& p) {
  out << '(' << p.first << ", " << p.second << ')';
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> const& v) {
  long long l = v.size();
  for (long long i = 0; i < l - 1; i++) out << v[i] << ' ';
  if (l > 0) out << v[l - 1];
  return out;
}
template <typename T>
void trace(const char* name, T&& arg1) {
  cout << name << " : " << arg1 << "\n";
}
template <typename T, typename... Args>
void trace(const char* names, T&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  trace(comma + 1, args...);
}
const long long N = 2 * 600;
const long long MOD = 998244353;
long long power(long long x, long long t) {
  long long ans = 1;
  while (t > 0) {
    if (t & 1) ans = 1LL * ans * x % MOD;
    x = 1LL * x * x % MOD;
    t >>= 1;
  }
  return ans;
}
inline long long sub(long long a, long long b) {
  a %= MOD;
  b %= MOD;
  a = ((a - b) % MOD + MOD) % MOD;
  return a;
}
inline long long add(long long a, long long b) {
  a %= MOD;
  b %= MOD;
  a = (a + b) % MOD;
  return a;
}
inline long long mul(long long a, long long b) {
  a %= MOD;
  b %= MOD;
  a = (a * b) % MOD;
  return a;
}
inline long long inv(long long a) {
  a = a % MOD;
  a = power(a, MOD - 2);
  return a;
}
long long m;
long long dp[N][N];
long long n;
vector<long long> t_arr, arr;
vector<vector<long long> > occ(N);
long long f(long long l, long long r) {
  if (l > r) return 1;
  if (dp[l][r] != -1) {
    return dp[l][r];
  }
  long long mi = 1000000000000000001, idx;
  for (long long i = l; i <= r; i++) {
    if (arr[i] < mi) {
      mi = arr[i];
      idx = i;
    }
  }
  if (l == r) {
    long long a = occ[mi][0];
    long long b = occ[mi].back();
    if (l > a || r < b) {
      return dp[l][r] = 0;
    }
    return dp[l][r] = 1;
  }
  long long a = occ[mi][0];
  long long b = occ[mi].back();
  if (l > a || r < b) {
    return dp[l][r] = 0;
  }
  long long ans_left = 0, ans_right = 0;
  for (long long rp = b; rp <= r; rp++) {
    if (arr[rp] >= arr[idx]) {
      ans_right = add(ans_right, mul(f(b + 1, rp), f(rp + 1, r)));
    } else {
      break;
    }
  }
  for (long long lp = a; lp >= l; lp--) {
    if (arr[lp] >= arr[idx]) {
      ans_left = add(ans_left, mul(f(l, lp - 1), f(lp, a - 1)));
    } else {
      break;
    }
  }
  long long ans = mul(ans_left, ans_right);
  for (long long i = 0; i < (long long)occ[mi].size() - 1; i++) {
    if (occ[mi][i] + 1 != occ[mi][i + 1]) {
      ans = mul(ans, f(occ[mi][i] + 1, occ[mi][i + 1] - 1));
    }
  }
  return dp[l][r] = ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m;
  t_arr.assign(m + 1, 0);
  for (long long i = 1; i <= m; i++) {
    cin >> t_arr[i];
  }
  for (long long i = 1; i <= m; i++) {
    if (arr.empty() || arr.back() != t_arr[i]) {
      arr.push_back(t_arr[i]);
    }
  }
  m = arr.size();
  if (m > 2 * n) {
    cout << 0 << "\n";
    return 0;
  }
  n = arr.size();
  for (long long i = 0; i < n; i++) {
    occ[arr[i]].push_back(i);
  }
  memset(dp, -1, sizeof(dp));
  cout << f(0, n - 1) << "\n";
}