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 N = 555;
long long dp[N][N], minc[N][N];
long long c[N], p[N];
long long n, m;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (long long i = (1); i <= (m); i++) {
    cin >> c[i];
    p[c[i]] = i;
  }
  for (long long l = (1); l <= (m); l++) {
    for (long long r = (l); r <= (m); r++) {
      if (l <= r) {
        minc[l][r] = n + 1;
        for (long long x = (l); x <= (r); x++) {
          minc[l][r] = min(minc[l][r], c[x]);
        }
      }
    }
  }
  for (long long l = (0); l <= (m + 1); l++) {
    for (long long r = (0); r <= (m + 1); r++) {
      dp[l][r] = 1;
    }
  }
  for (long long s = (2); s <= (m); s++) {
    for (long long l = (1); l <= (m - s + 1); l++) {
      long long r = l + s - 1;
      long long pos = p[minc[l][r]];
      long long x = 0, y = 0;
      for (long long i = (l); i <= (pos); i++) {
        x += dp[l][i - 1] * dp[i][pos - 1] % 998244353;
        x %= 998244353;
      }
      for (long long i = (pos); i <= (r); i++) {
        y += dp[pos + 1][i] * dp[i + 1][r] % 998244353;
        y %= 998244353;
      }
      dp[l][r] = x * y % 998244353;
    }
  }
  long long ans = dp[1][m];
  cout << ans << "\n";
  return 0;
}