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 S, class T>
pair<S, T> operator+(const pair<S, T> &s, const pair<S, T> &t) {
  return pair<S, T>(s.first + t.first, s.second + t.second);
}
template <class S, class T>
pair<S, T> operator-(const pair<S, T> &s, const pair<S, T> &t) {
  return pair<S, T>(s.first - t.first, s.second - t.second);
}
template <class S, class T>
ostream &operator<<(ostream &os, pair<S, T> p) {
  os << "(" << p.first << ", " << p.second << ")";
  return os;
}
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
const long long inf = 1000000001;
const long long INF = (long long)1e18 + 1;
const long long MOD = 998244353;
const double pi = 3.14159265358979323846;
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
int dx2[8] = {1, 1, 0, -1, -1, -1, 0, 1}, dy2[8] = {0, 1, 1, 1, 0, -1, -1, -1};
int n, m;
vector<int> c;
vector<vector<long long> > hyou(555, vector<long long>(555, -1));
long long f(int l, int r) {
  if (r == l) return 1;
  if (r - l == 1) return 1;
  if (hyou[l][r] >= 0) return hyou[l][r];
  int mi = inf;
  int idx = -1;
  for (int i = l; i < r; i++) {
    if (mi > c[i]) {
      mi = c[i];
      idx = i;
    }
  }
  long long temp = 0;
  for (int j = idx + 1; j <= r; j++) {
    long long temp2 = 1;
    if (j - (idx + 1) >= 2) {
      (temp2 *= f(idx + 1, j)) %= MOD;
    }
    if (r - j >= 2) {
      (temp2 *= f(j, r)) %= MOD;
    }
    temp = (temp + temp2) % MOD;
  }
  long long res = 0;
  for (int i = l; i <= idx; i++) {
    long long add = 1;
    if (i - l >= 2) {
      (add *= f(l, i)) %= MOD;
    }
    if (idx - i >= 2) {
      (add *= f(i, idx)) %= MOD;
    }
    res += temp * add;
    res %= MOD;
  }
  hyou[l][r] = res;
  return res;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  ;
  cin >> n >> m;
  assert(n == m);
  c.resize(n);
  for (int i = 0; i < (n); i++) cin >> c[i];
  long long ans = f(0, n) % MOD;
  ans = (ans + MOD) % MOD;
  cout << ans << endl;
}