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 maxn = 5e2 + 14, maxm = 1e6 + 14, mod = 998244353;
int n, m, mn[maxn], mx[maxn], a[maxm];
int dp[2 * maxn][2 * maxn];
vector<int> occ[maxn];
bool brk(int al, int ar, int bl, int br) {
  if (al > bl) swap(al, bl), swap(ar, br);
  return al < bl && bl < ar && ar < br;
}
int rem[maxm], add[maxm];
int solve(int l = 0, int r = m) {
  if (l + 1 >= r) return 1;
  if (dp[l][r] != -1) return dp[l][r];
  stack<int> st;
  for (int i = l; i < r; i++) {
    if (st.size() && st.top() > a[i]) return dp[l][r] = 0;
    if (add[i] != -1) st.push(a[i]);
    if (rem[i] != -1) {
      if (st.empty() || st.top() != rem[i]) return dp[l][r] = 0;
      st.pop();
    }
  }
  if (st.size()) return dp[l][r] = 0;
  int mid = n;
  for (int i = n - 1; i >= 0; i--) {
    if (l <= mn[i] && mx[i] < r) mid = i;
    if (brk(l, r, mn[i], mx[i] + 1)) return dp[l][r] = 0;
  }
  if (mid == n) return 0;
  int L = 0, R = 0, in = 1;
  for (int i = 1; i < occ[mid].size(); i++)
    in = (long long)in * solve(occ[mid][i - 1] + 1, occ[mid][i]) % mod;
  for (int i = 0; i < n; i++)
    if (mn[i] <= mn[mid] && l <= mn[i])
      L = (L + (long long)solve(l, mn[i]) * solve(mn[i], mn[mid])) % mod;
  for (int i = 0; i < n; i++)
    if (mx[mid] <= mx[i] && mx[i] < r)
      R = (R + (long long)solve(mx[mid] + 1, mx[i] + 1) * solve(mx[i] + 1, r)) %
          mod;
  return dp[l][r] = (long long)L * in % mod * R % mod;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  memset(mn, 63, sizeof mn);
  memset(dp, -1, sizeof dp);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> a[i];
    a[i]--;
  }
  m = unique(a, a + m) - a;
  if (m > 2 * n) return cout << "0\n", 0;
  for (int i = 0; i < m; i++) {
    int x = a[i];
    mn[x] = min(mn[x], i);
    mx[x] = i;
    occ[x].push_back(i);
  }
  memset(rem, -1, sizeof rem);
  memset(add, -1, sizeof rem);
  for (int i = 0; i < n; i++) {
    rem[mx[i]] = i;
    add[mn[i]] = i;
  }
  cout << solve() << '\n';
}