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;
const int MAXN = 505;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int c[MAXN];
long long int dp[MAXN][MAXN];
long long int get(int i, int j) {
  if (i >= j) return 1;
  return dp[i][j];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= m; i++) cin >> c[i];
  for (int i = m; i >= 1; i--)
    for (int j = 1; j <= m; j++) {
      if (i >= j) {
        dp[i][j] = 1;
        continue;
      }
      int mini = 1e9, id;
      for (int h = i; h <= j; h++)
        if (c[h] < mini) {
          mini = c[h];
          id = h;
        }
      long long int izq = get(i, id - 1), der = get(id + 1, j);
      for (int h = i; h < id; h++)
        izq += get(i, h) * get(h + 1, id - 1) % 998244353;
      izq %= 998244353;
      for (int h = id + 1; h <= j; h++)
        der += get(id + 1, h) * get(h + 1, j) % 998244353;
      der %= 998244353;
      dp[i][j] = izq * der % 998244353;
    }
  cout << dp[1][m] << '\n';
  return 0;
}