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;
class Solver {
 private:
  int n, m;
  vector<int> a;
  vector<int> lx, rx;
  struct seg {
    int x;
    int l;
    int r;
  };
  long long dfs(int l, int r, int p) {
    vector<seg> s;
    long long ans = 1;
    int now = l;
    while (now < r) {
      if (a[now] == p) {
        now++;
        int sz = s.size();
        if (sz > 0) {
          vector<vector<long long>> dp(sz + 1, vector<long long>(sz + 1, 0));
          for (int i = 0; i < (sz); i++) {
            if (s[i].r - s[i].l >= 3) {
              ans = ans * dfs(s[i].l + 1, s[i].r - 1, s[i].x) % 998244353;
            }
            dp[i][i + 1] = 1;
            dp[i][i] = 1;
          }
          dp[sz][sz] = 1;
          for (int w = (2); w < (sz + 1); w++) {
            for (int i = 0; i < (sz - w + 1); i++) {
              int l = i, r = i + w;
              int mi = 1000000000;
              int idx = -1;
              for (int k = (l); k < (r); k++) {
                if (mi > s[k].x) {
                  mi = s[k].x;
                  idx = k;
                }
              }
              long long suml = 0, sumr = 0;
              for (int x = (l); x < (idx + 1); x++) {
                suml = (suml + dp[l][x] * dp[x][idx] % 998244353) % 998244353;
              }
              for (int x = (idx + 1); x < (r + 1); x++) {
                sumr =
                    (sumr + dp[idx + 1][x] * dp[x][r] % 998244353) % 998244353;
              }
              dp[l][r] = suml * sumr % 998244353;
            }
          }
          ans = ans * dp[0][sz] % 998244353;
        }
        s.clear();
        continue;
      }
      int x = a[now];
      if (p > x) {
        cout << 0 << endl;
        exit(0);
      }
      s.push_back({x, lx[x], rx[x] + 1});
      now = rx[x] + 1;
    }
    int sz = s.size();
    if (sz > 0) {
      vector<vector<long long>> dp(sz + 1, vector<long long>(sz + 1, 0));
      for (int i = 0; i < (sz); i++) {
        if (s[i].r - s[i].l >= 3) {
          ans = ans * dfs(s[i].l + 1, s[i].r - 1, s[i].x) % 998244353;
        }
        dp[i][i + 1] = 1;
        dp[i][i] = 1;
      }
      dp[sz][sz] = 1;
      for (int w = (2); w < (sz + 1); w++) {
        for (int i = 0; i < (sz - w + 1); i++) {
          int l = i, r = i + w;
          int mi = 1000000000;
          int idx = -1;
          for (int k = (l); k < (r); k++) {
            if (mi > s[k].x) {
              mi = s[k].x;
              idx = k;
            }
          }
          long long suml = 0, sumr = 0;
          for (int x = (l); x < (idx + 1); x++) {
            suml = (suml + dp[l][x] * dp[x][idx] % 998244353) % 998244353;
          }
          for (int x = (idx + 1); x < (r + 1); x++) {
            sumr = (sumr + dp[idx + 1][x] * dp[x][r] % 998244353) % 998244353;
          }
          dp[l][r] = suml * sumr % 998244353;
        }
      }
      ans = ans * dp[0][sz] % 998244353;
    }
    return ans;
  }

 public:
  void solve() {
    cin >> n >> m;
    a = vector<int>(m);
    for (int i = 0; i < (m); i++) cin >> a[i], a[i]--;
    lx = vector<int>(n, -1);
    rx = vector<int>(n, -1);
    for (int i = 0; i < (m); i++) {
      if (lx[a[i]] == -1) lx[a[i]] = i;
      rx[a[i]] = i;
    }
    cout << dfs(0, m, -1) << endl;
  }
};
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  Solver s;
  s.solve();
  return 0;
}