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;
inline char gc() {
  static char now[1 << 16], *S, *T;
  if (T == S) {
    T = (S = now) + fread(now, 1, 1 << 16, stdin);
    if (T == S) return EOF;
  }
  return *S++;
}
inline long long read() {
  register long long x = 0, f = 1;
  char ch = gc();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = gc();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = gc();
  }
  return (f == 1) ? x : -x;
}
long long f[1010][1010], n, m, a[101010];
signed main() {
  n = read(), m = read();
  for (int i = 1; i <= m; i++) a[i] = read();
  for (int i = 1; i <= m; i++) f[i][i] = 1;
  for (int i = 0; i <= m + 1; i++) {
    for (int j = 0; j <= m + 1; j++) {
      f[i][j] = 1;
    }
  }
  for (int len = 2; len <= m; len++) {
    for (int i = 1; i + len - 1 <= m; i++) {
      int j = i + len - 1, x = 0;
      a[0] = n + 1;
      for (int k = i; k <= j; k++) {
        if (a[k] < a[x]) x = k;
      }
      long long tmp1 = f[i][x - 1], tmp2 = f[x + 1][j];
      for (int k = i; k < x; k++) {
        tmp1 = (tmp1 + f[i][k] * f[k + 1][x - 1] % 998244353) % 998244353;
      }
      for (int k = x + 1; k <= j; k++) {
        tmp2 = (tmp2 + f[x + 1][k] * f[k + 1][j] % 998244353) % 998244353;
      }
      f[i][j] = tmp1 * tmp2 % 998244353;
    }
  }
  cout << f[1][m];
  return 0;
}