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 <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  for (auto i : v) os << i << " ";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v) {
  for (auto i : v) os << i << " ";
  return os;
}
const long long MOD = 998244353;
const long long N = 1050;
long long a[N], f[N][N], idx[N][N];
long long dfs(long long l, long long r) {
  if (f[l][r] != -1) return f[l][r];
  long long ret = 0;
  long long pos = idx[l][r];
  long long p1 = 0, p2 = 0, p;
  for (long long L = l; L <= pos; ++L) {
    p = 1;
    if (L > l) p = 1ll * p * dfs(l, L - 1) % MOD;
    if (L < pos) p = 1ll * p * dfs(L, pos - 1) % MOD;
    p1 = (1ll * p1 + p) % MOD;
  }
  for (long long R = pos; R <= r; ++R) {
    p = 1;
    if (R > pos) p = 1ll * p * dfs(pos + 1, R) % MOD;
    if (R < r) p = 1ll * p * dfs(R + 1, r) % MOD;
    p2 = (1ll * p2 + p) % MOD;
  }
  return f[l][r] = (1ll * p1 * p2) % MOD;
}
int main() {
  memset(f, -1, sizeof(f));
  long long n, m;
  scanf("%I64d%I64d", &n, &m);
  for (long long i = 0; i < n; ++i) scanf("%I64d", &a[i]);
  for (long long i = 0; i < n; ++i) idx[i][i] = i;
  for (long long l = 1; l < n; ++l)
    for (long long i = 0; i + l - 1 < n; ++i) {
      long long j = i + l - 1;
      if (i > 1) idx[i - 1][j] = (a[i - 1] < a[idx[i][j]]) ? i - 1 : idx[i][j];
      if (j < n) idx[i][j + 1] = (a[j + 1] < a[idx[i][j]]) ? j + 1 : idx[i][j];
    }
  printf("%I64d\n", dfs(0, n - 1));
  return 0;
}