Problem: You are given a sequence of N integers: A_1,A_2,...,A_N.

Find the number of permutations p_1,p_2,...,p_N of 1,2,...,N that can be changed to A_1,A_2,...,A_N by performing the following operation some number of times (possibly zero), modulo 998244353:

* For each 1\leq i\leq N, let q_i=min(p_{i-1},p_{i}), where p_0=p_N. Replace the sequence p with the sequence q.

Constraints

* 1 \leq N \leq 3 × 10^5
* 1 \leq A_i \leq N
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
A_1
:
A_N


Output

Print the number of the sequences that satisfy the condition, modulo 998244353.

Examples

Input

3
1
2
1


Output

2


Input

5
3
1
4
1
5


Output

0


Input

8
4
4
4
1
1
1
2
2


Output

24


Input

6
1
1
6
2
2
2


Output

0

Solution (in CPP):

Solution:
#include <algorithm>
#include <cassert>
#include <cfloat>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <deque>
#include <iostream>
#include <limits>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#include <vector>
 
#define FOR(i,k,n) for (int (i)=(k); (i)<(n); ++(i))
#define rep(i,n) FOR(i,0,n)
#define all(v) begin(v), end(v)
#define debug(x) //std::cerr<< #x <<": "<<x<<"\n"
#define debug2(x,y) std::cerr<< #x <<": "<< x <<", "<< #y <<": "<< y <<"\n"
 
typedef long long ll;
typedef std::pair<int, int> pii;
typedef std::vector<int> vi;
typedef std::vector<std::vector<int> > vvi;
typedef std::vector<ll> vll;
typedef std::vector<std::vector<ll> > vvll;
typedef std::deque<bool> db;
template<class T> using vv=std::vector<std::vector< T > >;

// cout vector
template<typename T> std::ostream& operator<<(std::ostream& s, const std::vector<T>& v) {
  int len=v.size();s<<"\n";rep(i,len){s<<v[i];if(i<len-1){s << "\t";}}s<<"\n";return s; }
// cout 2-dimentional vector
template<typename T> std::ostream& operator<<(std::ostream& s, const std::vector<std::vector<T> >& vv) {
  int len=vv.size();rep(i,len){s<<vv[i];} return s; }
// cout deque
template<typename T> std::ostream& operator<<(std::ostream& s, const std::deque<T>& v) {
  int len=v.size();s<<"\n";rep(i,len){s<<v[i];if(i<len-1){s << "\t";}}s<<"\n";return s; }
// cout 2-dimentional deque
template<typename T> std::ostream& operator<<(std::ostream& s, const std::deque<std::deque<T> >& vv) {
  int len=vv.size();rep(i,len){s<<vv[i];} return s; }
// cout set<cout-able>
template<typename T> std::ostream& operator<<(std::ostream& s, const std::set<T>& v) {
  s<<"\n";for(auto elm : v){s<<elm<<"\t";}s<<"\n";return s; }

inline void scan(int&a){scanf("%d",&a);}
inline void scan(int&a,int&b){scanf("%d%d",&a,&b);}
inline void scan(int&a,int&b,int&c){scanf("%d%d%d",&a,&b,&c);}
inline void scan(int&a,int&b,int&c,int&d){scanf("%d%d%d%d",&a,&b,&c,&d);}
inline void scan(std::vector<int>&v){int sz=v.size();rep(i,sz){scan(v[i]);}}
inline void scan(std::vector<std::vector<int> >&v){int sz=v.size();rep(i,sz){scan(v[i]);}}
inline void scan(ll&a){scanf("%lld",&a);}
inline void scan(ll&a,ll&b){scanf("%lld%lld",&a,&b);}
inline void scan(ll&a,ll&b,ll&c){scanf("%lld%lld%lld",&a,&b,&c);}
inline void scan(ll&a,ll&b,ll&c,ll&d){scanf("%lld%lld%lld%lld",&a,&b,&c,&d);}
inline void scan(std::vector<ll>&v){int sz=v.size();rep(i,sz){scan(v[i]);}}
inline void scan(std::vector<std::vector<ll> >&v){int sz=v.size();rep(i,sz){scan(v[i]);}}
inline void scan(char&a){scanf(" %c",&a);}
inline void scan(std::vector<char>&v){int sz=v.size();rep(i,sz){scan(v[i]);}}
inline void scan(std::vector<std::vector<char> >&v){int sz=v.size();rep(i,sz){scan(v[i]);}}
inline void scan(std::string&s){char BUF[3000000];scanf(" %s",BUF);s=std::string(BUF);}
inline void scan(std::deque<int>&v){int sz=v.size();rep(i,sz){scan(v[i]);}}

inline void print(int a){printf("%d\n",a);}
inline void print(int a,int b){printf("%d %d\n",a,b);}
inline void print(ll a){printf("%lld\n",a);}
inline void print(ll a,ll b){printf("%lld %lld\n",a,b);}
inline void print(std::string s){std::cout << s << "\n";}
inline void print_yn(bool flg){if(flg){printf("Yes\n");}else{printf("No\n");}}

using namespace std;

int mod = 998244353;

int main() {
  int n;
  scan(n);
  deque<int> a(n);
  scan(a);
  rep (i, n+1) {
    if (i == n) {
      if (a[0] == 1) {
        ll ans = 1;
        rep (j, n) {
          ans = ans * (j + 1) % mod;
        }
        print(ans);
      } else {
        print(0);
      }
      return 0;
    }
    if (a[0] == a[n-1]) {
      a.push_front(a[n-1]);
      a.pop_back();
    } else {
      break;
    }
  }
  // 連続領域取得 重複検査を兼ねる
  int cur = -1;
  map<int, int> mp;
  rep (i, n) {
    if (cur != a[i]) {
      cur = a[i];
      if (mp.find(cur) != end(mp)) {
        print(0);
        return 0;
      }
      mp[cur] = i;
    }
  }

  deque<vector<int>> elms;
  rep (i, n) {
    int sz = elms.size();
    if (sz == 0 || elms[sz-1][0] != a[i]) {
      elms.push_back({a[i], 0});
    }
  }
  int m = elms.size();
  rep (i, m) {
    int tmpv = mp[elms[i][0]];
    elms[i][1] = tmpv;
  }
  map<int, int> index;
  rep (i, m) {
    index[elms[i][0]] = i + 1;
  }
  elms.push_back({elms[0][0], elms[0][1] + n});
  elms.push_front({elms[m-1][0], elms[m-1][1] - n});
  debug(elms);

  if (index.find(1) == end(index)) {
    print(0);
    return 0;
  }
  int k = elms[index[1] + 1][1] - elms[index[1]][1];

  ll ans = 1;
  int sum_space = 0;

  FOR (i, 1, n+1) {
    if (index.find(i) == end(index)) {
      if (sum_space == 0) {
        print(0);
        return 0;
      }
      ans = ans * sum_space % mod;
      sum_space -= 1;
    debug(ans);
    debug(sum_space);
      continue;
    }
    int ind = index[i];
    int prev_ind = ind - 1;
    int next_ind = ind + 1;
    int k_ = elms[next_ind][1] - elms[ind][1];
    if (k_ > k) {
      print(0);
      return 0;
    }
    if (elms[prev_ind][0] > i && i < elms[next_ind][0]) {
      if (k_ == k) {

      } else {
        print(0);
        return 0;
      }
    } else if (elms[prev_ind][0] > i && i > elms[next_ind][0]) {
      sum_space += (elms[next_ind][1] - elms[ind][1] - 1);
    } else if (elms[prev_ind][0] < i && i < elms[next_ind][0]) {
      sum_space += (k_ - 1);
    } else if (elms[prev_ind][0] < i && i > elms[next_ind][0]) {
      ans = ans * (elms[ind][1] - (elms[next_ind][1] - k) + 1) % mod;
      sum_space += (k-1 + k_ - 1);
    }
    debug(ans);
    debug(sum_space);
  }
  assert(sum_space == 0);
  print(ans);

  return 0;
}