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 long long MOD = 998244353;
vector<int> logn(513);
int main() {
  logn[1] = 0;
  for (int i = 2; i < 513; ++i) {
    logn[i] = logn[i / 2] + 1;
  }
  long long n, m;
  cin >> n >> m;
  vector<int> str(m);
  for (int i = 0; i < m; ++i) {
    cin >> str[i];
    str[i]--;
  }
  vector<vector<pair<int, int>>> go(logn[m] + 1, vector<pair<int, int>>(m));
  for (int i = 0; i < m; ++i) {
    go[0][i] = {str[i], i};
  }
  for (int lg = 1; lg <= logn[m]; ++lg) {
    for (int i = 0; i < m; ++i) {
      if (i + (1 << (lg - 1)) < m) {
        go[lg][i] = min(go[lg - 1][i], go[lg - 1][i + (1 << (lg - 1))]);
      } else {
        go[lg][i] = go[lg - 1][i];
      }
    }
  }
  vector<vector<long long>> dp(m + 1, vector<long long>(m + 1, 1));
  for (int len = 1; len <= m; ++len) {
    int lg = logn[len];
    for (int i = 0; i <= m - len; ++i) {
      auto p = min(go[lg][i], go[lg][i + len - (1 << lg)]);
      int ind = p.second;
      long long left = 0, right = 0;
      for (int j = i; j <= ind; ++j) {
        left += dp[j][ind] * dp[i][j];
        left %= MOD;
      }
      for (int j = ind + 1; j <= i + len; ++j) {
        right += dp[ind + 1][j] * dp[j][i + len];
        right %= MOD;
      }
      dp[i][i + len] = (left * right) % MOD;
    }
  }
  cout << dp[0][m] << endl;
  return 0;
}