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;
template <class T>
struct rge {
  T b, e;
};
template <class T>
rge<T> range(T i, T j) {
  return rge<T>{i, j};
}
struct debug {
  template <class T>
  debug& operator<<(const T&) {
    return *this;
  }
};
using ll = long long;
template <typename T>
void min_self(T& a, T b) {
  a = min(a, b);
}
template <typename T>
void max_self(T& a, T b) {
  a = max(a, b);
}
const int mod = 998244353;
inline int add(int a, int b) {
  a += b;
  while (a >= mod) a -= mod;
  while (a < 0) a += mod;
  return a;
}
inline void add_self(int& a, int b) { a = add(a, b); }
inline int mul(int a, int b) { return 1LL * a * b % mod; }
inline void mul_self(int& a, int b) { a = mul(a, b); }
inline int mpow(int a, int n) {
  int res = 1;
  while (n > 0) {
    if (n & 1) {
      res = mul(res, a);
    }
    a = mul(a, a);
    n >>= 1;
  }
  return res;
}
inline int divide(int a, int b) {
  assert(b != 0);
  return mul(a, mpow(b, mod - 2));
}
const int mxN = 200 * 1000 + 5;
int fact[mxN], finv[mxN];
inline int binomial(int n, int k) {
  return mul(fact[n], mul(finv[n - k], finv[k]));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<int> a(m);
  for (int i = 0; i < m; i++) {
    cin >> a[i];
    --a[i];
  }
  a.resize(unique(a.begin(), a.end()) - a.begin());
  m = a.size();
  if (m > 2 * n) {
    cout << "0\n";
    return 0;
  }
  vector<vector<int>> pos(n);
  for (int i = 0; i < m; i++) {
    pos[a[i]].push_back(i);
  }
  vector<vector<int>> dp(m + 1, vector<int>(m + 1, 1));
  for (int len = 1; len <= m; len++) {
    for (int i = 0; i <= m - len; i++) {
      int small = a[i];
      for (int j = 0; j < len; j++) {
        min_self(small, a[i + j]);
      }
      int L = pos[small][0] - i;
      int R = pos[small].back() - i;
      if (L < 0 || R >= len) {
        dp[i][len] = 0;
        continue;
      }
      int sum_on_left = 0, sum_on_right = 0;
      for (int j = 0; j <= L; j++) {
        add_self(sum_on_left, mul(dp[i][j], dp[i + j][L - j]));
      }
      for (int j = R + 1; j <= len; j++) {
        add_self(sum_on_right,
                 mul(dp[i + R + 1][j - (R + 1)], dp[i + j][len - j]));
      }
      dp[i][len] = mul(sum_on_left, sum_on_right);
      for (int j = 0; j < (int)pos[small].size() - 1; j++) {
        if (pos[small][j] + 1 != pos[small][j + 1]) {
          mul_self(
              dp[i][len],
              dp[pos[small][j] + 1][pos[small][j + 1] - pos[small][j] - 1]);
        }
      }
    }
  }
  cout << dp[0][m] << '\n';
}