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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod=1e9+7;
const ll INF=mod*mod;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
#define all(x) (x).begin(),(x).end()
ll min(ll a,ll b){
  if(a>b){
    return b;
    }
      else{
        return a;
      }
  }
  
  ll max(ll a,ll b){
    if(a>b){
      return a;
    }
    else{
      return b;
    }
  }

ll power(ll a,ll b){
 ll ans=1;
 for(ll i=1;i<=b;i++){
ans*=a;
  }
  return ans;
}

 ll gcd(ll m,ll n){
 	if(n==0){
       return m;
     }
   return gcd(n,m%n);
 }
ll lcm(ll a,ll b){
  ll g=gcd(a,b);
  return a/gcd(a,b)*b;
		

}
int main(){
  ll mod2=998244353;
  
  ll N;
  cin>>N;
  
  vector<ll>D(N);
  vector<ll>d(N);
 
  bool flag=false;
  rep(i,N){
    cin>>D[i];
    if(D[0]!=0){
      flag=true;
    }
 if(i!=0&&D[i]==0){
flag=true;
}
  }
  sort(all(D));
  
 for(ll i=1;i<N;i++){
if(D[i]!=D[i-1]){
if(D[i]-D[i-1]!=1){
flag=true;
}
}
 }

  ll num=0;
  ll ans=1;
  
  if(flag){
    cout<<0<<endl;
  }
else{
sort(all(D));

for(ll i=1;i<N;i++){
  if(D[i]!=D[i-1]){
    num=D[i-1];
    
  }
  
d[i]=(D.end()-lower_bound(all(D),num))-(D.end()-lower_bound(all(D),D[i]));
}
  
  for(ll i=1;i<N;i++){
    d[i]%=mod2;
ans*=d[i];
    ans%=mod2;
  }
  cout<<ans<<endl;

}
  

  
}