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;
unsigned long long min(unsigned long long x, unsigned long long y) {
  if (x < y) return x;
  return y;
}
unsigned long long max(unsigned long long x, unsigned long long y) {
  if (x < y) return y;
  return x;
}
long long min(long long x, long long y) {
  if (x < y) return x;
  return y;
}
long long max(long long x, long long y) {
  if (x < y) return y;
  return x;
}
double min(double x, double y) {
  if (x < y) return x;
  return y;
}
double max(double x, double y) {
  if (x < y) return y;
  return x;
}
unsigned long long gcd(unsigned long long x, unsigned long long y) {
  if (!x) return y;
  if (!y) return x;
  if (x > y) swap(x, y);
  return gcd(x, y % x);
}
unsigned long long inv(unsigned long long a, unsigned long long c) {
  if (a == 1) {
    return 1;
  }
  return ((c - (c / a)) * inv(c % a, c)) % c;
}
long long n, m, dp[505][505] = {0}, c[505] = {0}, M = 998244353;
long long mul(long long a, long long b) { return (a * b) % M; }
void add(long long &a, long long b) {
  a += b;
  if (a >= M) a -= M;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> c[i];
  }
  for (int size = 1; size <= m; size++) {
    for (int i = 0; i < m; i++) {
      int j = i + size - 1;
      if (j >= m) break;
      if (size == 1) {
        dp[i][j] = 1;
        continue;
      }
      int mine = INT_MAX, mindx = -1;
      for (int k = i; k <= j; k++) {
        if (c[k] < mine) {
          mine = c[k];
          mindx = k;
        }
      }
      if ((mindx > i) && (mindx < j)) {
        dp[i][j] = mul(dp[i][mindx], dp[mindx][j]);
        continue;
      }
      if (mindx == i) {
        for (int k = i; k <= j; k++) {
          if ((k > i) && (k < j))
            add(dp[i][j], mul(dp[i + 1][k], dp[k + 1][j]));
          else
            add(dp[i][j], dp[i + 1][j]);
        }
      } else {
        for (int k = j; k >= i; k--) {
          if ((k > i) && (k < j))
            add(dp[i][j], mul(dp[k][j - 1], dp[i][k - 1]));
          else
            add(dp[i][j], dp[i][j - 1]);
        }
      }
    }
  }
  cout << dp[0][m - 1] << "\n";
  return 0;
}