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 <class T>
T read1() {
  T t = 0;
  bool vis = 0;
  char k;
  do (k = getchar()) == '-' && (vis = 1);
  while ('0' > k || k > '9');
  while ('0' <= k && k <= '9')
    t = (t << 3) + (t << 1) + (k ^ '0'), k = getchar();
  return vis ? -t : t;
}
int dp[505][505], s, m, w[505][505], f[505][505], a[505];
int main() {
  s = read1<int>(), m = read1<int>();
  dp[0][0] = dp[s + 1][s + 1] = 1;
  for (int i = 1; i <= s; ++i)
    a[i] = read1<int>(), w[i][i] = i, dp[i][i] = 1, f[i][i] = 1;
  for (int i = 1; i < s; ++i)
    for (int j = i + 1; j <= s; ++j)
      w[i][j] = a[w[i][j - 1]] < a[j] ? w[i][j - 1] : j;
  for (int i = 1; i <= s + 1; ++i) f[i][i - 1] = dp[i][i - 1] = 1;
  for (int len = 2; len <= s; ++len)
    for (int i = 1, j; j = i + len - 1, i <= s - len + 1; ++i) {
      int n = w[i][j];
      f[i][j] = dp[i][j] = 1ll * dp[i][n - 1] * dp[n + 1][j] % 998244353;
      for (int k = i; k < j; ++k)
        dp[i][j] = (dp[i][j] + 1ll * dp[i][k] * f[k + 1][j]) % 998244353;
    }
  printf("%d", dp[1][s]);
  return 0;
}