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 <long long mod>
struct modular {
  long long value;
  modular(long long x = 0) {
    value = x % mod;
    if (value < 0) value += mod;
  }
  modular& operator+=(modular other) {
    if ((value += other.value) >= mod) value -= mod;
    return *this;
  }
  modular& operator-=(modular other) {
    if ((value -= other.value) < 0) value += mod;
    return *this;
  }
  modular& operator*=(modular other) {
    value = value * other.value % mod;
    return *this;
  }
  modular& operator/=(modular other) {
    long long a = 0, b = 1, c = other.value, m = mod;
    while (c != 0) {
      long long t = m / c;
      m -= t * c;
      swap(c, m);
      a -= t * b;
      swap(a, b);
    }
    a %= mod;
    if (a < 0) a += mod;
    value = value * a % mod;
    return *this;
  }
  modular operator-() { return modular(-value); }
  modular operator+(modular rhs) { return modular(*this) += rhs; }
  modular operator-(modular rhs) { return modular(*this) -= rhs; }
  modular operator*(modular rhs) { return modular(*this) *= rhs; }
  modular operator/(modular rhs) { return modular(*this) /= rhs; }
  bool operator==(modular rhs) { return value == rhs.value; }
  bool operator!=(modular rhs) { return value != rhs.value; }
  bool operator<(modular rhs) { return value < rhs.value; }
};
template <long long mod>
string to_string(modular<mod> x) {
  return to_string(x.value);
}
template <long long mod>
ostream& operator<<(ostream& stream, modular<mod> x) {
  x.value %= mod;
  if (x.value < 0) x.value += mod;
  return stream << x.value;
}
template <long long mod>
istream& operator>>(istream& stream, modular<mod>& x) {
  stream >> x.value;
  x.value %= mod;
  if (x.value < 0) x.value += mod;
  return stream;
}
const long long mod = 998244353LL;
using mint = modular<mod>;
inline mint pw(mint a, long long n) {
  mint res = 1;
  while (n > 0) {
    if (n & 1) {
      res *= a;
    }
    a *= a;
    n >>= 1;
  }
  return res;
}
vector<mint> fact(1, 1), finv(1, 1);
inline mint C(int n, int k) {
  if (n < k || k < 0) {
    return mint(0);
  }
  while ((int)fact.size() < n + 1) {
    fact.emplace_back(fact.back() * (int)fact.size());
    finv.emplace_back(mint(1) / fact.back());
  }
  return fact[n] * finv[k] * finv[n - k];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<int> a(m);
  for (int i = 0; i < m; i++) {
    cin >> a[i];
    a[i]--;
  }
  vector<vector<mint>> dp(n, vector<mint>(n));
  for (int l = n - 1; l >= 0; l--) {
    dp[l][l] = 1;
    for (int r = l + 1; r < n; r++) {
      int mn = n + 1, pos = -1;
      for (int p = l; p <= r; p++) {
        if (a[p] < mn) {
          mn = a[p];
          pos = p;
        }
      }
      mint x = 1;
      if (l < pos) {
        x = dp[l][pos - 1] * 2;
        for (int p = l; p + 1 < pos; p++) {
          x += dp[l][p] * dp[p + 1][pos - 1];
        }
      }
      mint y = 1;
      if (pos < r) {
        y = dp[pos + 1][r] * 2;
        for (int p = r; p - 1 > pos; p--) {
          y += dp[pos + 1][p - 1] * dp[p][r];
        }
      }
      dp[l][r] = x * y;
    }
  }
  cout << dp[0][n - 1] << '\n';
  return 0;
}