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>
#pragma GCC optimize("Ofast")
const int MOD = 998244353;
inline int lgput(int a, int b) {
  int ans = 1;
  while (b > 0) {
    if (b & 1) ans = (1LL * ans * a) % MOD;
    b >>= 1;
    a = (1LL * a * a) % MOD;
  }
  return ans;
}
inline void mod(int &x) {
  if (x >= MOD) x -= MOD;
}
inline void add(int &x, int y) {
  x += y;
  mod(x);
}
inline void sub(int &x, int y) {
  x += MOD - y;
  mod(x);
}
inline void mul(int &x, int y) { x = (1LL * x * y) % MOD; }
using namespace std;
const int MAXN = 505;
int dp[MAXN + 1][MAXN + 1], sp[MAXN + 1][MAXN + 1];
inline int get(int l, int r) {
  if (l > r) return 1;
  return dp[l][r];
}
int main() {
  int i, n, m;
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n >> m;
  vector<int> arr(n + 1);
  for (i = 1; i <= n; i++) {
    cin >> arr[i];
    dp[i][i] = sp[i + 1][i] = 1;
    sp[i][i] = 2;
  }
  for (int len = 1; len < n; len++) {
    for (int l = 1; l + len <= n; l++) {
      int r = l + len;
      int mn = 1e9, pos;
      for (int k = l; k <= r; k++) {
        if (mn > arr[k]) {
          mn = arr[k];
          pos = k;
        }
      }
      for (int a = pos; a >= l; a--) {
        dp[l][r] =
            (dp[l][r] + 1LL * (1LL * get(l, a - 1) * get(a, pos - 1) % MOD) *
                            sp[pos + 1][r]) %
            MOD;
      }
      for (int k = l - 1; k <= r; k++) {
        add(sp[l][r], (1LL * get(l, k) * get(k + 1, r)) % MOD);
      }
    }
  }
  cout << dp[1][n];
  return 0;
}