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 T, typename V>
ostream &operator<<(ostream &os, const pair<T, V> pai) {
  return os << '(' << pai.first << ' ' << pai.second << ')';
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> v) {
  cout << '[';
  for (auto p : v) cout << p << ",";
  cout << "]";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> v) {
  cout << "{";
  for (auto p : v) cout << p << ",";
  cout << "}";
  return os;
}
template <typename T, typename V>
ostream &operator<<(ostream &os, const map<T, V> v) {
  cout << "{";
  for (auto p : v) cout << p << ",";
  cout << "}";
  return os;
}
template <typename T>
void get_max(T &a, T b) {
  a = max(a, b);
}
template <typename T>
void get_min(T &a, T b) {
  a = min(a, b);
}
int N, M;
int A[1000001];
vector<int> place[501];
namespace SEG {
int N = 1 << 20;
int tree[1 << 21];
void init(int M, int *A) {
  for (int i = 0, _i = (M); i < _i; i++) tree[i + N] = A[i];
  for (int i = (N)-1, _i = (1); i >= _i; i--)
    tree[i] = min(tree[i << 1], tree[i << 1 | 1]);
}
int query(int l, int r) {
  int res = (1 << 30);
  l += N, r += N;
  while (l <= r) {
    if (l & 1) get_min(res, tree[l++]);
    if (~r & 1) get_min(res, tree[r--]);
    l >>= 1;
    r >>= 1;
  }
  return res;
}
}  // namespace SEG
void done() {
  cout << 0 << '\n';
  exit(0);
}
long long dp[1005][1005];
int main(void) {
  iostream::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  cin >> N >> M;
  int j = 1;
  for (int i = (1), _i = (M + 1); i < _i; i++) {
    int t;
    cin >> t;
    if (t != A[j - 1]) A[j] = t, place[t].push_back(j), j++;
  }
  M = j - 1;
  SEG::init(M + 1, A);
  place[0].push_back(0);
  place[0].push_back(M + 1);
  for (int i = (N + 1) - 1, _i = (1); i >= _i; i--) {
    int s = place[i][0], e = place[i][((int)(place[i]).size()) - 1];
    0;
    if (SEG::query(s, e) < i) done();
    vector<pair<long long, long long> > l, r;
    for (int j = e; A[j] >= i;
         j = place[A[j + 1]][((int)(place[A[j + 1]]).size()) - 1]) {
      long long res = (j == e ? 1 : 2 * dp[e + 1][j]);
      for (int k = e + 1; k < j;
           k = place[A[k + 1]][((int)(place[A[k + 1]]).size()) - 1])
        res += dp[e + 1][k] * dp[k + 1][j], res %= 998244353LL;
      r.push_back({j, res});
    }
    for (int j = s; A[j] >= i; j = place[A[j - 1]][0]) {
      long long res = (j == s ? 1 : 2 * dp[j][s - 1]);
      for (int k = e - 1; k > j; k = place[A[k - 1]][0])
        res += dp[j][k - 1] * dp[k][s - 1], res %= 998244353LL;
      l.push_back({j, res});
    }
    0;
    long long mult = 1;
    for (int j = 0, _j = (((int)(place[i]).size()) - 1); j < _j; j++)
      mult = mult * dp[place[i][j] + 1][place[i][j + 1] - 1] % 998244353LL;
    for (auto &i : (l))
      for (auto &j : (r))
        dp[i.first][j.first] =
            i.second * j.second % 998244353LL * mult % 998244353LL;
  }
  0;
  cout << dp[1][M] << '\n';
}