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>
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
using ll = long long;
using db = long double;
using ii = pair<int, int>;
const int N = 3e5 + 5, LG = 19, MOD = 998244353;
const int SQ = 225;
const long double EPS = 1e-7;
int dp[1005][1005];
int col[1005];
vector<int> go[505];
int solve(int i, int j) {
  if (i > j) return 1;
  int &ret = dp[i][j];
  if (~ret) return ret;
  int val = *min_element(col + i, col + j + 1);
  int lft = 0, rt = 0, idx;
  ret = 0;
  idx = go[val].back();
  if (idx > j) return ret;
  for (int k = idx; k < j + 1; k++)
    rt = (rt + solve(idx + 1, k) * 1ll * solve(k + 1, j)) % MOD;
  idx = go[val][0];
  if (idx < i) return ret;
  for (int k = i; k < idx + 1; k++)
    lft = (lft + solve(i, k - 1) * 1ll * solve(k, idx - 1)) % MOD;
  ret = 1ll * rt * lft % MOD;
  for (int k = 1; k < go[val].size(); k++)
    ret = 1ll * ret * solve(go[val][k - 1] + 1, go[val][k] - 1) % MOD;
  return ret;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  int ptr = 0;
  for (int _ = 0; _ < m; _++) {
    cin >> col[ptr++];
    if (ptr > 1 && col[ptr - 1] == col[ptr - 2]) --ptr;
    if (ptr > n + n) {
      cout << 0 << '\n';
      return 0;
    }
  }
  m = ptr;
  for (int i = 0; i < m; i++) go[col[i]].push_back(i);
  memset(dp, -1, sizeof dp);
  cout << solve(0, m - 1) << '\n';
  return 0;
}