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;
using ll = long long;
const ll MOD = 998244353;
const int N = 501;
ll dp[N][N];
ll solve(const vector<int>& col) {
  int n = col.size();
  if (n == 0) return 1;
  for (int i = 0; i < n; ++i) dp[i][i] = 1;
  for (int k = 2; k <= n; ++k) {
    for (int a = 0; a + k <= n; ++a) {
      int b = a + k - 1;
      int j = a;
      for (int i = a; i <= b; ++i) {
        if (col[i] < col[j]) j = i;
      }
      ll lefts = 0;
      for (int x = a; x <= j; ++x) {
        ll add = 1;
        if (x < j) add = (add * dp[x][j - 1]) % MOD;
        if (a < x) add = (add * dp[a][x - 1]) % MOD;
        lefts = (lefts + add) % MOD;
      }
      ll rights = 0;
      for (int y = j; y <= b; ++y) {
        ll add = 1;
        if (j < y) add = (add * dp[j + 1][y]) % MOD;
        if (y < b) add = (add * dp[y + 1][b]) % MOD;
        rights = (rights + add) % MOD;
      }
      dp[a][b] = (lefts * rights) % MOD;
    }
  }
  return dp[0][n - 1];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  ll res = 1;
  vector<int> sta;
  vector<bool> app(n, false);
  for (int i = 0; i < m; ++i) {
    int c;
    cin >> c;
    --c;
    if (app[c]) {
      vector<int> sub;
      while (!sta.empty() && sta.back() != c) {
        sub.push_back(sta.back());
        sta.pop_back();
      }
      if (sta.empty()) {
        res = 0;
      } else {
        for (auto v : sub) {
          if (v < c) res = 0;
        }
        res = (res * solve(sub)) % MOD;
      }
    } else {
      sta.push_back(c);
      app[c] = true;
    }
  }
  res = (res * solve(sta)) % MOD;
  cout << res << '\n';
}