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 maxm = 1000006;
const int maxn = 505;
const int mdn = 998244353;
int n, m, l;
int c[maxm];
int a[maxm];
long long dp[maxn << 1][maxn << 1];
vector<int> oc[maxn];
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    cin >> c[i];
  }
  for (int i = 1; i <= m; i++) {
    if (i == 1 || c[i] != c[i - 1]) a[++l] = c[i];
  }
  for (int i = 1; i <= l; i++) {
    oc[a[i]].push_back(i);
  }
  if (l > n + n) {
    cout << 0;
    return 0;
  }
  for (int i = 0; i <= l + 1; i++) {
    for (int j = 0; j <= l + 1; j++) {
      dp[i][j] = 1;
    }
  }
  for (int i = l; i >= 1; i--) {
    for (int j = i; j <= l; j++) {
      int mi = a[i];
      for (int k = i; k <= j; k++) {
        if (a[k] < mi) mi = a[k];
      }
      int fv = *oc[mi].begin(), ev = *(oc[mi].end() - 1);
      if (fv < i || ev > j)
        dp[i][j] = 0;
      else {
        long long mul = 1;
        for (int i = 1; i < oc[mi].size(); i++) {
          mul = (mul * (dp[oc[mi][i - 1] + 1][oc[mi][i] - 1])) % mdn;
        }
        long long sm0 = 0, sm1 = 0;
        for (int k = i; k <= fv; k++) {
          sm0 += dp[k][fv - 1] * dp[i][k - 1];
          sm0 %= mdn;
        }
        for (int k = j; k >= ev; k--) {
          sm1 += dp[ev + 1][k] * dp[k + 1][j];
          sm1 %= mdn;
        }
        cerr << fv << ',' << ev << endl;
        dp[i][j] = mul * sm0 % mdn * sm1 % mdn;
      }
    }
  }
  cout << dp[1][l];
  return 0;
}