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 long long MOD = 998244353;
template <class T>
inline bool MX(T &l, const T &r) {
  return l < r ? l = r, 1 : 0;
}
template <class T>
inline bool MN(T &l, const T &r) {
  return l > r ? l = r, 1 : 0;
}
int N;
int M;
int C[501];
int rev[501];
long long dp[2][502][502];
long long imos[502];
void Add(long long &d, long long v) {
  d += v;
  while (d >= MOD) d -= MOD;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(0);
  cin >> N >> M;
  for (int(i) = 0; (i) < (N); ++(i)) {
    cin >> C[i];
    C[i]--;
    rev[C[i]] = i;
  }
  int src = 0;
  int dst = 1;
  for (int(i) = 0; (i) < (N); ++(i)) {
    dp[src][i][i + 1] = 1;
    dp[src][i][i] = 1;
  }
  dp[src][N][N] = 1;
  vector<long long> L(N);
  vector<long long> R(N + 1);
  for (int(i) = (N)-1; (i) >= 0; --(i)) {
    for (int(l) = 0; (l) < (N); ++(l)) {
      L[l] = 0;
      for (int(p) = l; (p) < (rev[i] + 1); ++(p)) {
        Add(L[l], dp[src][l][p] * dp[src][p][rev[i]] % MOD);
      }
    }
    for (int(r) = 1; (r) < (N + 1); ++(r)) {
      R[r] = 0;
      for (int(q) = rev[i] + 1; (q) < (r + 1); ++(q)) {
        Add(R[r], dp[src][rev[i] + 1][q] * dp[src][q][r] % MOD);
      }
    }
    for (int(l) = 0; (l) < (N); ++(l)) {
      for (int(r) = l + 1; (r) < (N + 1); ++(r)) {
        if (rev[i] < l || r <= rev[i]) {
          dp[dst][l][r] = dp[src][l][r];
        } else {
          Add(dp[dst][l][r], L[l] * R[r] % MOD);
        }
      }
    }
    swap(src, dst);
    fill(dp[dst][0], dp[dst + 1][0], 0);
    for (int(j) = 0; (j) < (N + 1); ++(j)) {
      dp[src][j][j] = 1;
    }
  }
  cout << dp[src][0][N] << endl;
}