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>
#pragma GCC optimize("O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int arr = 2e5 + 10;
const int ar = 2e3 + 10;
const long double pi = acos(-1);
const long double eps = 1e-10;
const long long md = 1e9 + 7;
int a[1000000 + 10];
int a1[(int)(500 + 10)];
int a2[(int)(500 + 10)];
bool edge[(int)(500 + 10)][(int)(500 + 10)];
bool new_edge[(int)(500 + 10)][(int)(500 + 10)];
vector<int> reb[(int)(500 + 10)];
vector<vector<int>> dp[(int)(500 + 10)];
int min_in_subtree[(int)(500 + 10)];
int get_dp(int val, int l, int r) {
  if (l > r) {
    return 1;
  }
  if (reb[val].empty()) {
    return 1;
  }
  if (dp[val][l][r] != -1) {
    return dp[val][l][r];
  }
  int res = 0;
  int pos = l;
  for (int i = l; i <= r; i++) {
    if (reb[val][i] < reb[val][pos]) {
      pos = i;
    }
  }
  int sum1 = 0;
  for (int k = pos; k >= l; k--) {
    if (k <= pos - 1 && a2[reb[val][k]] != a1[reb[val][k + 1]] - 1) {
      break;
    }
    sum1 += 1ll * get_dp(val, l, k - 1) * get_dp(val, k, pos - 1) %
            (long long)(998244353ll);
    if (sum1 >= (long long)(998244353ll)) {
      sum1 -= (long long)(998244353ll);
    }
  }
  int sum2 = 0;
  for (int k = pos; k <= r; k++) {
    if (k >= pos + 1 && a1[reb[val][k]] != a2[reb[val][k - 1]] + 1) {
      break;
    }
    sum2 += 1ll * get_dp(val, pos + 1, k) * get_dp(val, k + 1, r) %
            (long long)(998244353ll);
    if (sum2 >= (long long)(998244353ll)) {
      sum2 -= (long long)(998244353ll);
    }
  }
  dp[val][l][r] =
      1ll * sum1 * sum2 % (long long)(998244353ll) *
      get_dp(reb[val][pos], 0, (int)(reb[reb[val][pos]].size()) - 1) %
      (long long)(998244353ll);
  return dp[val][l][r];
}
bool in(int a, int l, int r) { return l <= a && a <= r; }
bool in(int l1, int r1, int l2, int r2) {
  return in(l1, l2, r2) && in(r1, l2, r2);
}
int intersect(int l1, int r1, int l2, int r2) {
  return max(0, min(r1, r2) - max(l1, l2) + 1);
}
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    cin >> a[i];
    a2[a[i]] = i;
  }
  for (int i = m; i >= 1; i--) {
    a1[a[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    if (*min_element(a + a1[i], a + a2[i] + 1) < i) {
      cout << 0 << "\n";
      return 0;
    }
  }
  a1[0] = 0;
  a2[0] = m + 1;
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= n; j++) {
      if (i != j) {
        if (in(a1[j], a2[j], a1[i], a2[i])) {
          edge[i][j] = 1;
          new_edge[i][j] = 1;
          if (j < i) {
            cout << 0 << "\n";
            return 0;
          }
        } else if (!in(a1[i], a2[i], a1[j], a2[j]) &&
                   intersect(a1[j], a2[j], a1[i], a2[i]) != 0) {
          cout << 0 << "\n";
          return 0;
        }
      }
    }
  }
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= n; j++) {
      for (int k = 0; k <= n; k++) {
        if (edge[i][j] && edge[j][k]) {
          new_edge[i][k] = 0;
        }
      }
    }
  }
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= n; j++) {
      if (new_edge[i][j]) {
        reb[i].push_back(j);
      }
    }
  }
  for (int i = 0; i <= n; i++) {
    dp[i].assign(reb[i].size(), vector<int>(reb[i].size(), -1));
    sort(reb[i].begin(), reb[i].end(),
         [&](int l1, int l2) { return a1[l1] < a1[l2]; });
  }
  cout << get_dp(0, 0, (int)(reb[0].size()) - 1) << "\n";
}