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 mod = 998244353;
long long dp[510][510];
int t[510];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m, i, len, j, l, r;
  cin >> n >> m;
  for (i = 1; i <= n; i++) cin >> t[i];
  for (len = 1; len <= n; len++) {
    for (l = 1; l + len - 1 <= n; l++) {
      r = l + len - 1;
      if (len == 1) {
        dp[l][r] = 1;
        continue;
      }
      int mi = l;
      for (i = l; i <= r; i++)
        if (t[i] < t[mi]) mi = i;
      long long sl = dp[l][mi - 1] * 2 % mod;
      long long sr = dp[mi + 1][r] * 2 % mod;
      for (i = l; i + 1 < mi; i++)
        sl = (sl + dp[l][i] * dp[i + 1][mi - 1]) % mod;
      for (i = mi + 1; i <= r; i++)
        sr = (sr + dp[mi + 1][i] * dp[i + 1][r]) % mod;
      sl = max(sl, (long long)1);
      sr = max(sr, (long long)1);
      dp[l][r] = (sl * sr) % mod;
    }
  }
  cout << dp[1][n];
}