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;
const int mod = 998244353;
int n, m, t, c[1001], first[500], last[500];
long long dp[1000][1000];
long long f(int l, int r) {
  if (l > r) return 1;
  if (dp[l][r] != -1) return dp[l][r];
  int col = n;
  for (int i = l; i <= r; i++) col = min(col, c[i]);
  if (first[col] < l || r < last[col]) return 0;
  long long s1 = 0, s2 = 0;
  for (int a = l; a <= first[col]; a++)
    s1 = (s1 + f(l, a - 1) * f(a, first[col] - 1)) % mod;
  for (int b = last[col]; b <= r; b++)
    s2 = (s2 + f(last[col] + 1, b) * f(b + 1, r)) % mod;
  dp[l][r] = s1 * s2 % mod;
  int x = first[col];
  for (int i = first[col] + 1; i <= last[col]; i++) {
    if (c[i] != col) continue;
    dp[l][r] = (dp[l][r] * f(x + 1, i - 1)) % mod;
    x = i;
  }
  return dp[l][r];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> t;
  for (int i = 0; i < t; i++) {
    cin >> c[m];
    c[m]--;
    if (m == 0 || c[m - 1] != c[m]) m++;
    if (m > 2 * n) {
      cout << "0\n";
      return 0;
    }
  }
  for (int i = 0; i < m; i++) last[c[i]] = i;
  for (int i = m - 1; 0 <= i; i--) first[c[i]] = i;
  for (int i = 0; i < m; i++)
    for (int j = 0; j < m; j++) dp[i][j] = -1;
  cout << f(0, m - 1) << "\n";
  return 0;
}