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;
int gcd(int a, int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long modularExponentiation(long long base, long long power,
                                long long mod) {
  long long ans = 1;
  while (power > 0) {
    if (power & 1) ans = (ans * base) % mod;
    power = power >> 1;
    base = (base * base) % mod;
  }
  return ans;
}
const int N = 1010;
long long n, m, c, l[N], r[N], dp[N][N];
vector<long long> col, pos[N];
void solve() {
  cin >> n >> m;
  col.push_back(0);
  for (int i = 1; i <= m; ++i) {
    cin >> c;
    if (col.back() != c) col.push_back(c);
  }
  if ((int)col.size() - 1 > 2 * n) {
    cout << 0 << "\n";
    return;
  }
  for (int i = 1; i <= (int)col.size() - 1; ++i) {
    r[col[i]] = i;
    pos[col[i]].push_back(i);
  }
  for (int i = (int)col.size() - 1; i >= 1; --i) l[col[i]] = i;
  for (int i = 1; i <= (int)col.size(); ++i) {
    dp[i][i - 1] = 1;
    dp[i][i] = 1;
  }
  for (int len = 1; len <= (int)col.size() - 1; ++len) {
    for (int st = 1; st + len - 1 < (int)col.size(); ++st) {
      int en = st + len - 1;
      long long minC = INT_MAX;
      for (int i = st; i <= en; ++i) minC = min(minC, col[i]);
      if (l[minC] < st || r[minC] > en) {
        dp[st][en] = 0;
        continue;
      }
      long long lans = 0, rans = 0, mans = 0;
      for (int i = st; i <= l[minC]; ++i)
        lans =
            (lans + dp[st][i - 1] * dp[i][l[minC] - 1] % 998244353) % 998244353;
      for (int i = r[minC]; i <= en; ++i)
        rans =
            (rans + dp[r[minC] + 1][i] * dp[i + 1][en] % 998244353) % 998244353;
      dp[st][en] = lans * rans % 998244353;
      for (int i = 0; i <= (int)pos[minC].size() - 2; ++i)
        dp[st][en] = (dp[st][en] * dp[pos[minC][i] + 1][pos[minC][i + 1] - 1]) %
                     998244353;
    }
  }
  cout << dp[1][(int)col.size() - 1] << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}