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 int inf = 0x3f3f3f3f;
const long long __64inf = 0x3f3f3f3f3f3f3f3f;
const int MAX = 87000;
const int Mod = 998244353;
template <class T>
struct F {
  T x;
  F(T x = 0) : x(x) {}
  inline F &operator+=(const F &other) {
    x = (x + other.x) % Mod;
    if (x < 0) x += Mod;
    return *this;
  }
  inline F &operator*=(const F &other) {
    x = (x * other.x) % Mod;
    if (x < 0) x += Mod;
    return *this;
  }
  inline F operator*(const F &other) const {
    T xx = x * other.x % Mod;
    if (xx < 0) xx += Mod;
    return F(xx);
  }
};
template <class T>
ostream &operator<<(ostream &out, const F<T> &x) {
  cout << x.x;
  return out;
}
template <class T>
F<T> operator*(long long a, const F<T> &b) {
  return F<T>((a * b.x) % Mod);
}
template <class T>
F<T> operator+(long long a, const F<T> &b) {
  return F<T>((a + b.x) % Mod);
}
F<long long> dp[505][505];
int main() {
  int n, m;
  cin >> n >> m;
  vector<int> col(m + 1);
  for (int i = 1; i <= m; i++) {
    cin >> col[i];
  }
  for (int i = 0; i <= n + 1; i++) {
    for (int j = 0; j <= n + 1; j++) {
      if (i >= j) dp[i][j] = 1;
    }
  }
  for (int L = 1; L < n; L++) {
    for (int i = 1; i + L <= n; i++) {
      pair<int, int> lo = {col[i], i};
      for (int j = i; j <= L + i; j++) lo = min(lo, {col[j], j});
      int pos = lo.second;
      F<long long> cnt1, cnt2;
      for (int a = i; a <= pos; a++) {
        cnt1 += dp[i][a - 1] * dp[a][pos - 1];
      }
      for (int b = pos; b <= i + L; b++) {
        cnt2 += dp[pos + 1][b] * dp[b + 1][i + L];
      }
      dp[i][i + L] = cnt1 * cnt2;
    }
  }
  cout << dp[1][n] << endl;
  return 0;
}