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;
long long dp[1010][1010];
long long komm[1010][1010];
int n, m;
vector<int> t[501];
vector<int> v;
long long mod = 998244353;
long long solve(int a, int b);
long long kom(int a, int b) {
  if (komm[a][b] != 0) {
    return komm[a][b];
  }
  long long ans = 0;
  for (int i = a; i <= b; i++) {
    ans += solve(a, i - 1) * solve(i, b - 1);
    ans %= mod;
  }
  komm[a][b] = ans;
  return ans;
}
long long solve(int a, int b) {
  if (dp[a][b] != 0) {
    return dp[a][b];
  }
  if (a >= b) {
    dp[a][b] = 1;
    return 1;
  }
  int pos = v[a];
  for (int i = a; i <= b; i++) {
    if (v[i] < pos) {
      pos = v[i];
    }
  }
  long long sum = 1;
  long long ans = 0;
  int x = t[pos][0];
  int y = t[pos].back();
  ans = kom(a, x);
  sum *= ans;
  ans = 0;
  ans = kom(y + 1, b + 1);
  sum *= ans;
  sum %= mod;
  ans = 0;
  for (int i = 1; i < t[pos].size(); i++) {
    x = t[pos][i - 1];
    y = t[pos][i];
    ans = solve(x + 1, y - 1);
    sum *= ans;
    sum %= mod;
  }
  dp[a][b] = sum;
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  v.push_back(0);
  for (int i = 1; i <= m; i++) {
    int x;
    cin >> x;
    if (x != v.back()) {
      v.push_back(x);
    }
  }
  m = v.size() - 1;
  for (int i = 1; i <= m; i++) {
    t[v[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = t[i][0]; j < t[i].back(); j++) {
      if (v[j] < i) {
        cout << 0;
        return 0;
      }
    }
  }
  solve(1, m);
  cout << dp[1][m] << endl;
  return 0;
}