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;
using ll = long long;
using ii = pair<int, int>;
using vi = vector<int>;
using vll = vector<ll>;
using vii = vector<ii>;
const ll MOD = 998244353;
const int INF = 1e9 + 9;
const int MAXM = 1000006;
const int MAXN = 502;
struct op {
  ii operator()(ii a, ii b) {
    if (a.second < b.second)
      return a;
    else
      return b;
  }
};
template <class T, class Operation = plus<T>>
struct SegmentTree {
  int n;
  vector<T> st;
  Operation op;
  T identity;
  T query(int L, int R) {
    T resl = identity, resr = identity;
    for (L += n, R += n; L < R; L >>= 1, R >>= 1) {
      if (L & 1) resl = op(resl, st[L++]);
      if (R & 1) resr = op(st[--R], resr);
    }
    return op(resl, resr);
  }
  void update(int p, T val) {
    for (st[p += n] = val; p > 1; p >>= 1) st[p >> 1] = op(st[p], st[p ^ 1]);
  }
  SegmentTree(const vector<T> &A, Operation op = Operation(),
              const T &identity = 0)
      : n((int)A.size()), st(n << 1), op(op), identity(identity) {
    copy(A.begin(), A.end(), st.begin() + n);
    for (int i = n - 1; i > 0; --i)
      st[i] = this->op(st[i << 1], st[i << 1 | 1]);
  }
};
int n, m, k, arr[MAXM], good, bad;
int l[MAXN], r[MAXN];
int LL[MAXN], RR[MAXN];
ll mem[MAXN][MAXN];
bool feasible = true, discard[MAXN];
SegmentTree<ii, op> st(vii(), op(), ii(INF, INF));
ll solve(int L, int R) {
  if (R < L) return 0;
  if (R - L <= 1) return 1;
  if (mem[L][R] != -1) return mem[L][R];
  int pos = st.query(L, R).first;
  ll suml = 0, sumr = 0;
  for (int i = L; i <= pos; i++) {
    suml = (suml + (solve(L, i) * solve(i, pos)) % MOD) % MOD;
  }
  for (int i = pos + 1; i <= R; i++) {
    sumr = (sumr + (solve(pos + 1, i) * solve(i, R)) % MOD) % MOD;
  }
  return mem[L][R] = (suml * sumr) % MOD;
}
bool comp(int a, int b) { return l[a] < l[b]; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  memset(mem, -1, sizeof(mem));
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    LL[i] = -1;
    l[i] = 2 * m;
    r[i] = -1;
    RR[i] = m;
  }
  for (int i = 0; i < m; i++) {
    cin >> arr[i];
    arr[i]--;
    r[arr[i]] = max(r[arr[i]], i);
    l[arr[i]] = min(l[arr[i]], i);
  }
  vii col;
  for (int i = n - 1; i >= 0; i--) {
    for (int j = 0; j < i; j++) {
      if (l[j] > r[i]) {
      } else if (r[j] < l[i]) {
      } else if (r[j] > r[i] && l[j] < l[i]) {
        discard[i] = true;
      } else {
        feasible = false;
      }
    }
    if (!discard[i]) col.push_back(ii(l[i], i));
  }
  sort(col.begin(), col.end());
  for (int i = 0; i < col.size(); i++) {
    col[i].first = i;
  }
  st = SegmentTree<ii, op>(col, op(), ii(INF, INF));
  if (feasible) {
    cout << solve(0, (int)col.size()) << endl;
  } else
    cout << 0 << endl;
}