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;
long long mod = 1000000007LL;
long long large = 2000000000000000000LL;
vector<int> compress(vector<int> a) {
  vector<int> t(1, a[0]);
  for (int i = 1; i < (int)a.size(); i++) {
    if (a[i] != a[i - 1]) t.push_back(a[i]);
  }
  return t;
}
vector<vector<long long> > dp;
vector<int> c, lb, rb;
int n, m;
long long f(int l, int r) {
  if (l > r) return 1;
  if (dp[l][r] != -1) return dp[l][r];
  int id = n;
  for (int i = l; i <= r; i++) id = min(id, c[i]);
  if (lb[id] < l || rb[id] > r) return dp[l][r] = 0;
  vector<int> t;
  for (int i = lb[id]; i <= rb[id]; i++) {
    if (c[i] == id) t.push_back(i);
  }
  long long off = 1;
  for (int i = 0; i + 1 < (int)t.size(); i++) {
    off *= f(t[i] + 1, t[i + 1] - 1);
    off %= mod;
  }
  long long le = 0, ri = 0;
  for (int i = lb[id]; i >= l; i--) {
    long long temp = 1;
    if (i <= lb[id] - 1) temp *= f(i, lb[id] - 1);
    if (i - 1 >= l) temp *= f(l, i - 1);
    temp %= mod;
    le += temp;
  }
  for (int i = rb[id]; i <= r; i++) {
    long long temp = 1;
    if (i >= rb[id] + 1) temp *= f(rb[id] + 1, i);
    if (i + 1 <= r) temp *= f(i + 1, r);
    temp %= mod;
    ri += temp;
  }
  ri %= mod;
  le %= mod;
  return dp[l][r] = off * le % mod * ri % mod;
}
int main() {
  mod = 998244353LL;
  cin >> n >> m;
  c.assign(m, 0);
  for (int i = 0; i < m; i++) {
    cin >> c[i];
    c[i]--;
  }
  c = compress(c);
  m = (int)c.size();
  if (m > 2 * n) {
    cout << 0 << endl;
    return 0;
  }
  lb.assign(n, m);
  rb.assign(n, -1);
  for (int i = 0; i < m; i++) {
    lb[c[i]] = min(lb[c[i]], i);
    rb[c[i]] = max(rb[c[i]], i);
  }
  dp.assign(m, vector<long long>(m, -1));
  cout << f(0, m - 1) << endl;
  return 0;
}