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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long n, m;
  cin >> n >> m;
  vector<long long> a((size_t)m);
  for (long long i = 0; i < m; i++) {
    cin >> a[i];
    a[i]--;
  }
  for (long long i = 0, j = 0; i < m; i++) {
    while (j != 0 && j < m && a[j - 1] == a[j]) {
      j++;
    }
    if (j == m) {
      a.erase(a.begin() + i, a.end());
      break;
    } else {
      a[i] = a[j];
      j++;
    }
  }
  m = a.size();
  if (m > n * 2) {
    cout << 0 << endl;
    return 0;
  }
  vector<pair<long long, long long>> c((size_t)n,
                                       {(long long)1e14, (long long)-1e14});
  for (long long i = 0; i < m; i++) {
    c[a[i]].first = min(c[a[i]].first, i);
    c[a[i]].second = max(c[a[i]].second, i);
  }
  vector<vector<long long>> b((size_t)m, vector<long long>((size_t)m));
  for (long long k = 0; k < m; k++) {
    for (long long i = 0; i + k < m; i++) {
      long long j = i + k;
      long long g = *min_element(a.begin() + i, a.begin() + j + 1);
      if (c[g].first < i || c[g].second > j) {
        b[i][j] = 0;
      } else if (j - i + 1 == 1) {
        b[i][j] = 1;
      } else {
        vector<long long> v;
        for (long long h = i; h <= j; h++) {
          if (a[h] == g) {
            v.push_back(h);
          }
        }
        long long x = 0, y = 0;
        for (long long h = i; h <= v.front(); h++) {
          x += (i > h - 1 ? 1 : b[i][h - 1]) *
               (h > v.front() - 1 ? 1 : b[h][v.front() - 1]);
          x %= 998244353;
        }
        for (long long h = j; h >= v.back(); h--) {
          y += (v.back() + 1 > h ? 1 : b[v.back() + 1][h]) *
               (h + 1 > j ? 1 : b[h + 1][j]);
          y %= 998244353;
        }
        b[i][j] = x * y;
        b[i][j] %= 998244353;
        for (long long h = 1; h < (long long)v.size(); h++) {
          b[i][j] *= v[h - 1] + 1 > v[h] - 1 ? 0 : b[v[h - 1] + 1][v[h] - 1];
          b[i][j] %= 998244353;
        }
      }
    }
  }
  cout << b[0][m - 1] << endl;
  return 0;
}