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>
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
using namespace std;
const int N = 505;
const int M = 1000 * 1000 + 5;
const int inf = 1000 * 1000 * 1000;
const int mod = 998244353;
int n, m;
int a[M];
vector<int> g[N];
int mp[3 * N][3 * N];
int fst[N], lst[N];
int t[4 * M];
void build(int v, int s, int e) {
  if (s == e) {
    t[v] = a[s];
    return;
  }
  int m = (s + e) / 2;
  build(2 * v, s, m);
  build(2 * v + 1, m + 1, e);
  t[v] = min(t[2 * v], t[2 * v + 1]);
}
int get(int v, int s, int e, int l, int r) {
  if (l > r) return inf;
  if (s == l && e == r) return t[v];
  int m = (s + e) / 2;
  return min(get(2 * v, s, m, l, min(r, m)),
             get(2 * v + 1, m + 1, e, max(m + 1, l), r));
}
int solve(int l, int r) {
  if (l == r + 1) return 1;
  if (mp[l][r]) return mp[l][r];
  int mn = get(1, 1, m, l, r);
  int pat = 1;
  for (int i = 1; i < (int)g[mn].size(); i++) {
    pat = pat * 1ll * solve(g[mn][i - 1] + 1, g[mn][i] - 1) % mod;
  }
  int s1 = solve(l, g[mn][0] - 1);
  vector<pair<int, int> > A;
  for (int i = 1; i <= n; i++) {
    if (g[i].back() >= g[mn][0] || g[i].back() < l) continue;
    A.push_back(make_pair(g[i].back(), i));
    A.push_back(make_pair(g[i][0], -i));
  }
  sort(A.begin(), A.end());
  set<int> q;
  for (int i = (int)A.size() - 1; i >= 0; i--) {
    if (A[i].second > 0) {
      q.insert(A[i].second);
    } else {
      q.erase(-A[i].second);
    }
    if ((int)q.size() == 0) {
      s1 += solve(l, A[i].first - 1) * 1ll * solve(A[i].first, g[mn][0] - 1) %
            mod;
      s1 %= mod;
    }
  }
  int s2 = solve(g[mn].back() + 1, r);
  A.clear();
  for (int i = 1; i <= n; i++) {
    if (g[i][0] <= g[mn].back() || g[i][0] > r) continue;
    A.push_back(make_pair(g[i][0], -i));
    A.push_back(make_pair(g[i].back(), i));
  }
  sort(A.begin(), A.end());
  for (int i = 0; i < (int)A.size(); i++) {
    if (A[i].second < 0) {
      q.insert(-A[i].second);
    } else {
      q.erase(A[i].second);
    }
    if ((int)q.size() == 0) {
      s2 += solve(A[i].first + 1, r) * 1ll *
            solve(g[mn].back() + 1, A[i].first) % mod;
      s2 %= mod;
    }
  }
  s1 = s1 * 1ll * s2 % mod;
  pat = pat * 1ll * s1 % mod;
  return mp[l][r] = pat;
}
int main() {
  mt19937 myrand(chrono::steady_clock::now().time_since_epoch().count());
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    scanf("%d", &a[i]);
  }
  int z = 0;
  for (int i = 1; i <= m; i++) {
    int x = a[i];
    while (i <= m && a[i] == x) {
      ++i;
    }
    --i;
    a[++z] = x;
  }
  m = z;
  for (int i = 1; i <= m; i++) {
    g[a[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j < i; j++) {
      int id = lower_bound(g[j].begin(), g[j].end(), g[i][0]) - g[j].begin();
      if (id == (int)g[j].size()) continue;
      if (g[j][id] > g[i].back()) continue;
      cout << 0 << endl;
      return 0;
    }
  }
  build(1, 1, m);
  cout << solve(1, m) << endl;
  return 0;
}