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;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
void pre() {}
void solve() {}
long long dp[509][509];
vector<int> a;
long long mod = 998244353;
long long f(int i, int j) {
  if (i > j) return 1ll;
  if (dp[i][j] >= 0) return dp[i][j];
  return -1;
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  pre();
  int n;
  cin >> n >> n;
  a.resize(n, 0);
  for (int i = 0; i < (n); ++i) cin >> a[i];
  for (int i = 0; i < (n + 1); ++i)
    for (int j = 0; j < (n + 1); ++j) dp[i][j] = -1;
  for (int i = 0; i < (n + 1); ++i) dp[i][i] = 1, dp[i + 1][i] = 1;
  for (int l = 1; l <= (n); ++l)
    for (int i = 0; i < (n); ++i)
      if (i + l <= n) {
        int j = i + l - 1;
        int mn = min_element(a.begin() + i, a.begin() + j + 1) - a.begin();
        long long l = 0, r = 0;
        for (int k = i; k <= (mn); ++k)
          l += f(i, k - 1) * f(k, mn - 1) % mod, l %= mod;
        for (int k = mn; k <= (j); ++k)
          r += f(mn + 1, k) * f(k + 1, j) % mod, r %= mod;
        dp[i][j] = l * r % mod;
      }
  cout << dp[0][n - 1] << endl;
  return 0;
}