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>
#pragma GCC optimize("O3")
using namespace std;
template <class T>
inline bool setmin(T &a, T b) {
  if (a > b) return a = b, 1;
  return 0;
}
template <class T>
inline bool setmax(T &a, T b) {
  if (a < b) return a = b, 1;
  return 0;
}
template <class T>
inline T fast(T a, T b, T mod) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
template <class T>
inline T russian(T a, T b, T mod) {
  long long res = 0;
  while (b) {
    if (b & 1) res = (res + a) % mod;
    a = (a + a) % mod;
    b >>= 1;
  }
  return res;
}
template <class T>
inline T sub(T a, T b, T mod) {
  return ((a % mod - b % mod) + mod) % mod;
}
template <class T>
inline T add(T a, T b, T mod) {
  return (a % mod + b % mod) % mod;
}
template <class T>
inline T mult(T a, T b, T mod) {
  return ((a % mod * b % mod) + mod) % mod;
}
template <class T>
inline T mult(vector<T> vec, T mod) {
  T ret = 1;
  for (int i = 0; i < (int)(vec).size(); i++) ret = (ret * vec[i]) % mod;
  return ret;
}
template <class T>
istream &operator>>(istream &os, vector<T> &container) {
  for (auto &u : container) os >> u;
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const vector<T> &container) {
  for (auto &u : container) os << u << " ";
  return os;
}
template <typename T>
inline T gcd(T a, T b) {
  while (b) swap(a %= b, b);
  return a;
}
long double e = 2.7182818284590452353602874713526624;
long double PI = acos(-1);
long double eps = 1e-11;
const long long T = 1 << 14;
const long long INF = 1e9 + 7;
const long long MOD = 998244353;
const long long BIG_INF = 1e18 + 7;
const long long N = 1e6 + 7;
const long long M = 2007;
long long n, m, k, t;
string w;
int tab[501];
long long mapa[501][501];
int find_min(int pocz, int kon) {
  int ret = -1;
  int mini = n + 1;
  for (int i = (pocz); i <= (kon); i++) {
    if (tab[i] < mini) ret = i;
    setmin(mini, tab[i]);
  }
  return ret;
}
long long rec(int pocz, int kon) {
  if (pocz >= kon) return 1;
  if (mapa[pocz][kon] != 0) return mapa[pocz][kon];
  int poz = find_min(pocz, kon);
  long long ret = 0;
  long long lewo = 0;
  for (int i = (pocz); i <= (poz); i++)
    lewo = (lewo + rec(pocz, i - 1) * rec(i, poz - 1)) % MOD;
  for (int j = (poz); j <= (kon); j++)
    ret = (ret + (rec(poz + 1, j) * rec(j + 1, kon) % MOD) * lewo) % MOD;
  mapa[pocz][kon] = ret;
  return ret;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m;
  for (int i = 0; i < (n); i++) cin >> tab[i];
  cout << rec(0, n - 1);
}