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 SZ = 500 + 10;
const long long MOD = 998244353;
int n, m;
long long arr[SZ];
long long dp[SZ][SZ];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  int r, ind, tl, tr;
  long long fir, sec;
  long long calc[4];
  for (int len = 1; len <= n; len++) {
    for (int l = 0; l <= (n - len); l++) {
      if (len == 1) {
        dp[l][l] = 1ll;
        continue;
      }
      r = l + len - 1;
      ind = l;
      for (int i = l; i <= r; i++) {
        if (arr[i] < arr[ind]) ind = i;
      }
      dp[l][r] = 0;
      fir = 0;
      sec = 0;
      for (tl = l; tl <= ind; tl++) {
        if (tl > (ind - 1))
          calc[0] = 1ll;
        else
          calc[0] = dp[tl][ind - 1];
        if (l > (tl - 1))
          calc[1] = 1ll;
        else
          calc[1] = dp[l][tl - 1];
        calc[0] *= calc[1];
        calc[0] %= MOD;
        fir += calc[0];
        fir %= MOD;
      }
      for (tr = ind; tr <= r; tr++) {
        if ((ind + 1) > tr)
          calc[2] = 1ll;
        else
          calc[2] = dp[ind + 1][tr];
        if ((tr + 1) > r)
          calc[3] = 1ll;
        else
          calc[3] = dp[tr + 1][r];
        calc[2] *= calc[3];
        calc[2] %= MOD;
        sec += calc[2];
        sec %= MOD;
      }
      dp[l][r] += fir * sec;
      dp[l][r] %= MOD;
    }
  }
  cout << dp[0][n - 1] << endl;
  return 0;
}