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>
#include <bits/stdc++.h>
const long double PI = acos(-1.0);
using namespace std;
int n, m;
long long dp[505][505];
int a[505];
long long cal(int l, int r) {
  if (l >= r) return 1;
  long long &ret = dp[l][r];
  if (ret != -1) return ret;
  ret = 0;
  int mn = n + 1, idx = 0;
  for (int i = l; i <= r; i++) {
    if (a[i] < mn) {
      mn = a[i];
      idx = i;
    }
  }
  long long lft = cal(l, idx - 1);
  long long rgh = cal(idx + 1, r);
  for (int i = l; i < idx; i++) {
    lft += cal(l, i) * cal(i + 1, idx - 1);
    lft %= 998244353;
  }
  for (int i = idx + 1; i <= r; i++) {
    rgh += cal(idx + 1, i) * cal(i + 1, r);
    rgh %= 998244353;
  }
  lft %= 998244353;
  rgh %= 998244353;
  ret = rgh * lft;
  ret %= 998244353;
  return ret;
}
void init() { memset(dp, -1, sizeof dp); }
void read_input() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> a[i];
}
void solve() {
  init();
  cal(0, n - 1);
}
void print_ans() { cout << cal(0, n - 1) << '\n'; }
int main() {
  int t = 1;
  while (t--) {
    read_input();
    solve();
    print_ans();
    if (t) init();
  }
}
