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 long long maxn = 1e3 + 10;
const long long mod = 998244353;
const long long base = 1e18;
long long a[maxn * maxn];
long long dp[maxn][maxn];
long long l[maxn];
long long r[maxn];
vector<long long> adj[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  if (fopen("t.inp", "r")) {
    freopen("test.inp", "r", stdin);
    freopen("test.out", "w", stdout);
  }
  long long n, m;
  cin >> n >> m;
  long long cnt = 0;
  for (int i = 1; i <= m; i++) {
    long long x;
    cin >> x;
    if (cnt == 0 || x != a[cnt]) {
      cnt++;
      a[cnt] = x;
    }
  }
  m = cnt;
  if (m > 2 * n) {
    cout << 0;
    return 0;
  }
  for (int i = 1; i <= m; i++) {
    adj[a[i]].push_back(i);
  }
  for (int i = 1; i <= m; i++) dp[i + 1][i] = 1;
  dp[1][0] = 1;
  for (int len = 1; len <= m; len++) {
    for (int i = 1; i + len - 1 <= m; i++) {
      long long j = i + len - 1;
      long long val = base;
      for (int k = i; k <= j; k++) {
        val = min(val, a[k]);
      }
      long long l = adj[val][0];
      long long r = adj[val].back();
      if (i <= l && j >= r) {
        long long left = 0;
        long long right = 0;
        for (int t = i; t <= l; t++) {
          left = (left + dp[i][t - 1] * dp[t][l - 1]) % mod;
        }
        for (int t = r; t <= j; t++) {
          right = (right + dp[r + 1][t] * dp[t + 1][j]) % mod;
        }
        dp[i][j] = (left * right) % mod;
        for (int t = 0; t < adj[val].size() - 1; t++) {
          dp[i][j] =
              (dp[i][j] * dp[adj[val][t] + 1][adj[val][t + 1] - 1]) % mod;
        }
      } else {
        dp[i][j] = 0;
      }
    }
  }
  cout << dp[1][m];
}