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 f[1001000], a[1001000];
long long dp[510][510];
vector<int> v[510];
int find(int x) {
  if (f[x] == x) return x;
  return f[x] = find(f[x]);
}
void dp_reset(int n) {
  for (int i = 0; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      dp[i][j] = -1;
    }
  }
}
long long solve(vector<int>& seq, int l, int r) {
  if (l == r) return 1;
  if (dp[l][r] >= 0) return dp[l][r];
  int mn = l;
  long long ans = 0;
  for (int i = l; i < r; i++) {
    if (seq[i] < seq[mn]) mn = i;
    ans += solve(seq, l, mn) * solve(seq, mn + 1, i + 1) % 998244353 *
           solve(seq, i + 1, r) % 998244353;
  }
  return dp[l][r] = ans % 998244353;
}
int main() {
  int n, m;
  long long ans = 1;
  bool gg = false;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d", &a[i]);
    v[a[i]].push_back(i);
    f[i] = i;
  }
  f[0] = 0;
  f[m + 1] = m + 1;
  v[0].push_back(0);
  v[0].push_back(m + 1);
  for (int i = n; i >= 0 && !gg; i--) {
    for (int j = 0; j + 1 < v[i].size() && !gg; j++) {
      if (v[i][j] + 1 == v[i][j + 1]) {
        f[v[i][j]] = v[i][j] + 1;
        continue;
      }
      int u = v[i][j];
      int cnt = -1;
      vector<int> seq;
      do {
        if (a[u + 1] < i) {
          gg = true;
          break;
        }
        f[u] = u + 1;
        cnt++;
        u = find(u);
        if (a[u] != i) seq.push_back(a[u]);
      } while (u != v[i][j + 1]);
      dp_reset(seq.size());
      ans = ans * solve(seq, 0, seq.size()) % 998244353;
    }
  }
  if (gg)
    puts("0");
  else
    printf("%lld\n", ans);
  return 0;
}