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;
int n, m;
int a[1000005];
long long dp[1005 + 5][1005 + 5];
vector<int> v[505];
void chmin(int &x, int y) {
  if (x > y) x = y;
}
void Add(long long &x, long long y) {
  x += y;
  if (x >= 998244353) x -= 998244353;
}
long long dfs(int l, int r) {
  if (l > r) return 1;
  if (l == r) {
    if (v[a[l]].size() <= 1) return 1;
    return 0;
  }
  if (dp[l][r] != -1) return dp[l][r];
  long long &cur = dp[l][r];
  int mn = 1e9;
  for (int i = l; i <= r; i++) mn = min(mn, a[i]);
  vector<int> w = v[mn];
  if (w.back() > r || w[0] < l) return 0;
  long long A = 0, B = 0;
  for (int i = l; i <= w[0]; i++)
    Add(A, dfs(l, i - 1) * dfs(i, w[0] - 1) % 998244353);
  for (int i = w.back(); i <= r; i++)
    Add(B, dfs(w.back() + 1, i) * dfs(i + 1, r) % 998244353);
  cur = A * B % 998244353;
  for (int i = 0; i < w.size() - 1; i++)
    cur = cur * dfs(w[i] + 1, w[i + 1] - 1) % 998244353;
  return cur;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    cin >> a[i];
    if (a[i] == a[i - 1]) i--, m--;
  }
  memset(dp, -1, sizeof(dp));
  if (m > 1005) return cout << "0\n", 0;
  for (int i = 1; i <= m; i++) v[a[i]].push_back(i);
  cout << dfs(1, m) << '\n';
  return 0;
}