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 base = 998244353;
const int maxn = 1005;
bool memo[maxn][maxn];
int f[maxn][maxn], rmq[maxn][maxn];
int c[maxn];
vector<int> adj[maxn];
int calc(int l, int r) {
  if (l > r) return 1;
  if (memo[l][r]) return f[l][r];
  memo[l][r] = 1;
  int pos = rmq[l][r];
  int First = *adj[c[pos]].begin(), Last = adj[c[pos]].back();
  if (First < l || Last > r) return 0;
  int64_t Left = 0;
  int64_t Right = 0;
  int64_t Mid = 1;
  for (int k = l; k <= First; ++k)
    Left = (Left + 1ll * calc(l, k - 1) * calc(k, First - 1)) % base;
  for (int k = Last; k <= r; ++k)
    Right = (Right + 1ll * calc(Last + 1, k) * calc(k + 1, r)) % base;
  int pre = First;
  for (int x : adj[c[pos]]) {
    Mid = Mid * calc(pre + 1, x - 1) % base;
    pre = x;
  }
  Mid %= base;
  return f[l][r] = Left * Mid % base * Right % base;
}
int main() {
  clock_t STARTTIME = clock();
  iostream::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  int len = n;
  n = 0;
  int pre = 0;
  for (int i = 1; i <= m; ++i) {
    int x;
    cin >> x;
    if (x != pre) {
      c[++n] = x;
      adj[x].push_back(n);
    }
    pre = x;
  }
  if (n >= len * 2) return !(cout << 0);
  for (int l = 1; l <= n; ++l) {
    int best = l;
    for (int r = l; r <= n; ++r) {
      if (c[r] < c[best]) best = r;
      rmq[l][r] = best;
    }
  }
  cout << calc(1, n) << endl;
  return 0;
}