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 long long INF = 200000000000000000LL;
const long double EPS = 1e-9;
const int inf = 0x3f3f3f3f;
const long long MOD = 998244353LL;
const long long MOD2 = MOD * MOD;
const int N = 507;
int a[N];
long long dp[N][N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= m; ++i) cin >> a[i];
  for (int i = 0; i < N; ++i) {
    for (int j = 0; j <= i; ++j) {
      dp[i][j] = 1;
    }
  }
  for (int len = 2; len <= n; ++len) {
    for (int l = 1, r = len; r <= n; ++l, ++r) {
      int pos = -1, minval = inf;
      for (int i = l; i <= r; ++i)
        if (a[i] < minval) minval = a[i], pos = i;
      long long sum1 = 0;
      for (int i = l; i <= pos; ++i) {
        sum1 += dp[l][i - 1] * dp[i][pos - 1];
        if (sum1 >= MOD2) sum1 -= MOD2;
      }
      sum1 %= MOD;
      long long sum2 = 0;
      for (int i = pos; i <= r; ++i) {
        sum2 += dp[pos + 1][i] * dp[i + 1][r];
        if (sum2 >= MOD2) sum2 -= MOD2;
      }
      sum2 %= MOD;
      dp[l][r] = (sum1 * sum2) % MOD;
    }
  }
  long long res = dp[1][n];
  cout << res << "\n";
  return 0;
}