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;
template <unsigned int N>
class Field {
 public:
  inline Field(int x = 0) : v(x) {}
  inline Field<N> &operator+=(const Field<N> &o) {
    if (v + o.v >= N)
      v += o.v - N;
    else
      v += o.v;
    return *this;
  }
  inline Field<N> &operator*=(const Field<N> &o) {
    v = (unsigned long long)v * o.v % N;
    return *this;
  }
  inline Field<N> operator*(const Field<N> &o) const {
    Field<N> r{*this};
    return r *= o;
  }
  inline explicit operator unsigned int() const { return v; }

 private:
  unsigned int v;
};
template <unsigned int N>
ostream &operator<<(std::ostream &os, const Field<N> &f) {
  return os << (unsigned int)f;
}
template <unsigned int N>
Field<N> operator*(int i, const Field<N> &f) {
  return Field<N>(i) * f;
}
int main() {
  ios_base::sync_with_stdio(false);
  int N, M;
  cin >> N >> M;
  vector<int> C(N);
  for (int &c : C) {
    cin >> c;
    --c;
  }
  vector<vector<Field<998244353>>> D(N + 1, vector<Field<998244353>>(N + 1, 1));
  for (int l = 1; l <= M; ++l) {
    for (int a = 0; a + l <= M; ++a) {
      pair<int, int> lo = {C[a], 0};
      for (int i = 0; i < l; ++i) lo = min(lo, {C[a + i], i});
      int j = lo.second;
      if (j < 0 || j >= l) {
        D[a][l] = 0;
        continue;
      }
      Field<998244353> left = 0, right = 0;
      for (int u = 0; u <= j; ++u) left += D[a][u] * D[a + u][j - u];
      for (int v = j + 1; v <= l; ++v)
        right += D[a + j + 1][v - (j + 1)] * D[a + v][l - v];
      D[a][l] = left * right;
    }
  }
  cout << D[0][N] << endl;
}