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 fun() {}
long long int __gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return __gcd(b, a % b);
}
long long int poww(long long int a, long long int b, long long int md) {
  if (b < 0) return 0;
  if (a == 0) return 0;
  long long int res = 1;
  while (b) {
    if (b & 1) {
      res = (1ll * res * a) % md;
    }
    a = (1ll * a * a) % md;
    b >>= 1;
  }
  return res;
}
long long int divide(long long int a, long long int b, long long int md) {
  long long int rr = a * (poww(b, md - 2, md));
  rr %= md;
  return rr;
}
const long long int size = 505;
long long int n, m;
long long int md = 998244353;
long long int c[size];
long long int dp[size][size];
long long int smallest[size][size];
long long int solve(long long int start, long long int end) {
  if (start >= end) return 1;
  if (dp[start][end] != -1) return dp[start][end];
  long long int pos = smallest[start][end];
  long long int rightans = 0;
  for (long long int b = pos; b <= end; b++) {
    rightans = (rightans + solve(pos + 1, b) * solve(b + 1, end) % md) % md;
  }
  long long int leftans = 0;
  for (long long int a = start; a <= pos; a++) {
    leftans = (leftans + solve(start, a - 1) * solve(a, pos - 1) % md) % md;
  }
  return dp[start][end] = rightans * leftans % md;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  fun();
  cin >> n >> m;
  for (long long int i = 1; i <= n; i++) {
    cin >> c[i];
  }
  for (long long int i = 1; i <= n; i++) {
    smallest[i][i] = i;
    for (long long int j = i + 1; j <= n; j++) {
      long long int sm = smallest[i][j - 1];
      if (c[j] < c[sm]) {
        smallest[i][j] = j;
      } else {
        smallest[i][j] = sm;
      }
    }
  }
  memset(dp, -1, sizeof(dp));
  cout << solve(1, n) << '\n';
  return 0;
}