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>
bool local = false;
using namespace std;
template <class T>
string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
template <class T>
void print(T v) {
  cout << ((int)(v).size()) << "\n";
  for (auto x : v) cout << x << ' ';
  cout << "\n";
};
template <class T1, class T2>
ostream &operator<<(ostream &o, pair<T1, T2> x) {
  return o << x.first << ' ' << x.second;
}
template <class T1, class T2>
istream &operator>>(istream &o, pair<T1, T2> &x) {
  return o >> x.first >> x.second;
}
template <class T1, class T2>
pair<T1, T2> operator+(pair<T1, T2> a, pair<T1, T2> b) {
  a.first += b.first;
  a.second += b.second;
  return a;
}
template <class T1, class T2>
pair<T1, T2> operator-(pair<T1, T2> a, pair<T1, T2> b) {
  a.first -= b.first;
  a.second -= b.second;
  return a;
}
template <class T1, class T2>
void operator+=(pair<T1, T2> &a, pair<T1, T2> b) {
  a.first += b.first;
  a.second += b.second;
}
template <class T1, class T2>
void operator-=(pair<T1, T2> &a, pair<T1, T2> b) {
  a.first -= b.first;
  a.second -= b.second;
}
int nint() {
  int x;
  cin >> x;
  return x;
}
double getTime() { return clock() / (double)CLOCKS_PER_SEC; };
mt19937 rnd(0);
int rand(int n) { return rnd() % n; }
int rand(int l, int r) { return rnd() % (r - l + 1) + l; }
const int mod = 998244353;
void initIO() {
  if (local) {
    freopen("input.txt", "r", stdin);
    srand((int)time(0));
    rnd.seed((int)time(0));
  } else {
  }
  {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  };
}
void solve();
int main() {
  initIO();
  int tc = 1;
  for (int tt = 0; tt < (tc); tt++) {
    solve();
  }
  if (local)
    cout << "\n"
         << "time = " << getTime() << "\n";
}
int n;
int m;
int pos[505];
int a[505];
int was[505];
long long getans(int x) {
  x = pos[x];
  int l = x, r = x;
  while (l >= 0 && !was[l]) l--;
  while (r < n && !was[r]) r++;
  return (long long)(x - l) * (r - x) % mod;
}
int table[505][505];
int getans(int l, int r) {
  if (l >= r) return 1;
  int &ans = table[l][r];
  if (ans != -1) return ans;
  ans = 0;
  int p = min_element(a + l, a + r + 1) - a;
  long long tmp1 = 0;
  long long tmp2 = 0;
  for (int p1 = l; p1 <= p; p1++) {
    long long tmp = (long long)getans(l, p1 - 1) * getans(p1, p - 1) % mod;
    tmp1 += tmp;
  }
  for (int p2 = p; p2 <= r; p2++) {
    long long tmp = (long long)getans(p + 1, p2) * getans(p2 + 1, r) % mod;
    tmp2 += tmp;
  }
  tmp1 %= mod;
  tmp2 %= mod;
  ans = tmp1 * tmp2 % mod;
  return ans;
}
void solve() {
  cin >> n >> m;
  for (int i = 0; i < (n); i++) {
    int x;
    cin >> x;
    x--;
    a[i] = x;
    pos[x] = i;
  }
  memset(table, -1, sizeof(table));
  cout << getans(0, n - 1) << "\n";
  return;
  long long ans = 1;
  for (int i = 0; i < (n); i++) {
    ans *= getans(i);
    ans %= mod;
    was[pos[i]] = 1;
  }
  cout << ans;
}