Problem:
Score : 300 points

Problem Statement

Given is an integer sequence D_1,...,D_N of N elements. Find the number, modulo 998244353, of trees with N vertices numbered 1 to N that satisfy the following condition:

For every integer i from 1 to N, the distance between Vertex 1 and Vertex i is D_i.

Notes

A tree of N vertices is a connected undirected graph with N vertices and N-1 edges, and the distance between two vertices are the number of edges in the shortest path between them.

Two trees are considered different if and only if there are two vertices x and y such that there is an edge between x and y in one of those trees and not in the other.

Constraints

1 \leq N \leq 10^5

0 \leq D_i \leq N-1

Input

Input is given from Standard Input in the following format:

N
D_1 D_2 ... D_N

Output

Print the answer.

Sample Input 1

4
0 1 1 2

Sample Output 1

2

For example, a tree with edges (1,2), (1,3), and (2,4) satisfies the condition.

Sample Input 2

4
1 1 1 1

Sample Output 2

0

Sample Input 3

7
0 3 2 1 2 2 1

Sample Output 3

24

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

//FIRST THINK THEN CODE.

#include <bits/stdc++.h>
 
 
using namespace std;
 
typedef long long ll;
 
//#define MOD 998244353
#define rep(i,a,b) for(ll i=a;i<b;++i)
#define rrep(i,a,b) for(ll i=a;i>b;--i)
#define FOR(i,n)  for(ll i=0;i<n;i++)
#define vi vector<int>
#define vl vector<ll>
#define ld long double
#define vvi vector<vector<int>>
#define vvl vector<vector<long long>>
#define pii pair<int,int>
#define pll pair<long,long>
#define vpii vector<pii>
#define vpll vector<pll>
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define bs binary_search
#define d1(x) cout<<(x)<<endl
#define d2(x,y) cout<<(x)<<" "<<(y)<<endl
#define d3(x,y,z) cout<<(x)<<" "<<(y)<<" "<<(z)<<endl
#define d4(a,b,c,d) cout<<(a)<<" "<<(b)<<" "<<(c)<<" "<<(d)<<endl
#define PI 3.1415926535897932384626433832795
#define fix(f,n) fixed<<setprecision(n)<<f
#define all(x) x.begin(),x.end()
#define rev(p) reverse(p.begin(),p.end());
#define endl "\n"
#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define popcount(x) __builtin_popcountll(x)
#define sz(x) ((ll)x.size())
const ll M=1000000007;
const ll MM=998244353;
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";

 
ll newmod(ll a,ll b)
 {
  return ((a%b)+b)%b;
}
 
 
ll powM(ll a,ll b,ll m )
{ 
  if(a<=0)return 0;
  a%=m;
  if(!b)return 1;
  ll ans=1LL;
  while(b)
  {
    if(b&1)ans=ans*a%m;
    a=a*a%m;
    b>>=1;
  }
 
return ans;
}


 
ll poww(ll a,ll b)
{ 
  
  ll ans=1;
  while(b)
  {
    if(b&1)ans=ans*a;
    a=a*a;
    b>>=1;
  }
 
return ans;

}

template<typename T,typename F>
void chmax( T &a,F b){
  if(b>a)a=b;
}

template<typename T,typename F>
void chmin( T &a,F b){
  if(b<a)a=b;
}

const ll N=1e5+5;



int main()
{ 

IOS;


ll n;cin>>n;

vl x(N);

ll mx=0;

for(int i=0;i<n;i++){
  ll d;cin>>d;
  if(!i&&d){
    cout<<0;
    return 0;
  }
  x[d]++;
  mx=max(mx,d);
}


if(x[0]!=1){
  cout<<0;
  return 0;
}

ll ans=1;

for(int i=1;i<=mx;i++){

ans=(ans*powM(x[i-1],x[i],MM))%MM;
if(!ans)break;
}

cout<<ans;

//end_routine();

return 0;
 
}