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 long long rd() {
  long long x = 0;
  bool neg = false;
  char c = getchar();
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  if (c == '-') neg = true, c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  if (neg) x = -x;
  return x;
}
inline void write(long long x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
const int N = 500;
int n, m, c[N + 2], dp[N + 2][N + 2];
const int mod = 998244353;
void input() {
  cin >> n >> m;
  for (int _b = m, i = 1; i <= _b; ++i) cin >> c[i];
}
int dfs(int l, int r) {
  if (dp[l][r]) return dp[l][r];
  if (l >= r) return dp[l][r] = 1;
  int p = l;
  for (int _b = r, i = l; i <= _b; ++i)
    if (c[i] < c[p]) p = i;
  int suml = 0, sumr = 0;
  for (int _b = p, i = l; i <= _b; ++i)
    suml = (suml + 1ll * dfs(l, i - 1) * dfs(i, p - 1) % mod) % mod;
  for (int _b = r, i = p; i <= _b; ++i)
    sumr = (sumr + 1ll * dfs(p + 1, i) * dfs(i + 1, r) % mod) % mod;
  return dp[l][r] = 1ll * suml * sumr % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  input();
  cout << dfs(1, m);
  return 0;
}