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 N = 1005, M = 1e6 + 5, MOD = 998244353;
int a[N], b[N], lb[N], rb[N];
bool used[N];
long long dp[N][N];
vector<int> occ[N];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<int> v;
  int last = -1, len = 0;
  for (int i = 0; i < m; i++) {
    int x;
    cin >> x;
    if (used[x]) {
      cout << 0;
      return 0;
    }
    if (last != x) {
      last = x;
      a[++len] = x;
    }
    while (!v.empty() && v.back() > x) {
      used[v.back()] = 1;
      v.pop_back();
    }
    if (v.empty() || v.back() != x) v.push_back(x);
  }
  for (int i = 1; i <= len; i++) {
    occ[a[i]].push_back(i);
  }
  for (int i = 1; i <= len + 1; i++) {
    dp[i][i - 1] = 1;
  }
  occ[0] = {0, len + 1};
  for (int i = n; i > 0; i--) {
    long long res = 1;
    for (int j = 1; j < occ[i].size(); j++) {
      res = res * dp[occ[i][j - 1] + 1][occ[i][j] - 1] % MOD;
    }
    for (int x = occ[i].front(); x > 0;) {
      for (int y = occ[i].back(); y <= len;) {
        long long s1 = 0, s2 = 0;
        for (int p = occ[i].front(); p >= x;) {
          s1 = (s1 + dp[x][p - 1] * dp[p][occ[i].front() - 1]) % MOD;
          p = occ[a[p - 1]].front();
        }
        for (int p = occ[i].back(); p <= y;) {
          s2 = (s2 + dp[occ[i].back() + 1][p] * dp[p + 1][y]) % MOD;
          p = occ[a[p + 1]].back();
        }
        dp[x][y] = s1 * s2 % MOD * res % MOD;
        y++;
        if (a[y] < i) break;
        occ[a[y]].back();
      }
      x--;
      if (a[x] < i) break;
      x = occ[a[x]].front();
    }
  }
  cout << dp[1][len];
  return 0;
}