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 double PI = acos(-1.0);
const int INF = 1000 * 1000 * 1000 + 7;
const long long LINF = (long long)INF * INF;
const int mod = 998244353;
const int MAX = 1 << 9;
void ADD(int& x, int val) {
  x += val;
  if (x >= mod) x -= mod;
}
int mult(int a, int b) { return (long long)a * b % mod; }
int c[MAX];
int mn[MAX][MAX];
int dp[MAX][MAX];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = (0); i < (m); ++i) cin >> c[i];
  for (int l = (0); l < (m); ++l) {
    mn[l][l] = l;
    for (int r = (l + 1); r < (m); ++r)
      mn[l][r] = c[r] < c[mn[l][r - 1]] ? r : mn[l][r - 1];
  }
  for (int i = (0); i < (m + 1); ++i) dp[i + 1][i] = 1;
  for (int len = (1); len < (m + 1); ++len) {
    for (int l = (0); l < (m + 1 - len); ++l) {
      int r = l + len - 1;
      int s1 = 0, s2 = 0;
      for (int i = (l); i < (mn[l][r] + 1); ++i)
        ADD(s1, mult(dp[l + 1][i], dp[i + 1][mn[l][r]]));
      for (int i = (mn[l][r]); i < (r + 1); ++i)
        ADD(s2, mult(dp[mn[l][r] + 2][i + 1], dp[i + 2][r + 1]));
      dp[l + 1][r + 1] = mult(s1, s2);
    }
  }
  cout << dp[1][m] << "\n";
  return 0;
}