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: Wrong Answer)

/*Function Template*/
#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef pair<ll,ll> pint;
const int MAX = 510000;
const int MOD = 998244353;
#define rep(i, n) for(ll i = 0; i < (n); i++)
#define Rep(i, n) for(ll i = 1; i < (n); i++)
#define ALL(a)  (a).begin(),(a).end()
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
ll fac[MAX], finv[MAX], inv[MAX];


template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (ll i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }

}
ll Len(ll n) {
  ll s=0;
  while(n!=0) s++, n/=10;
  return s;
}


ll Sint(ll n) {
  ll m=0,s=0,a=n;
  while(a!=0) s++, a/=10;
  for(ll i=s-1;i>=0;i--) m+=n/((ll)pow(10,i))-(n/((ll)pow(10,i+1)))*10;
  return m;
}


ll Svec(vector<ll> v){
  ll n=0;
  for(ll i=0;i<v.size();i++) n+=v[i];
  return n;
}


ll GCD(ll a,ll b) {
  return b ? GCD(b,a%b) : a;
}


ll LCM(ll a,ll b){
  return a/GCD(a,b)*b;
}


ll Factorial(ll n){
  ll m=1;
  while(n>=1) m*=n,n--;
  return m;
}

void runlength(string s,vector<pair<char,ll>> &p){
  ll x=1;
  if(s.size()==1){
    p.push_back(pair<char,ll>(s[0],1));
  }
  for(ll i=0;i<s.size()-1;i++){
    if(s[i]==s[i+1]){
      x++;
      if(i==s.size()-2){
        p.push_back(pair<char,ll>(s[i],x));
      }
    }else{
      p.push_back(pair<char,ll>(s[i],x));
      x=1;
      if(i==s.size()-2){
        p.push_back(pair<char,ll>(s[s.size()-1],x));
      }
    }
  }
}




ll COM(ll n,ll k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

const int MAX_N=100010;
vector<bool> sieve_of_eratosthenes(){
  vector<bool> isPrime(MAX_N+1,true);
  for(int i=2;i<=MAX_N;i++){
    if(isPrime[i]){
      for(int j=2*i;j<=MAX_N;j+=i){
        isPrime[j]=false;
      }
    }
  }
  return isPrime;
}

vector<pint> prime_factorize(ll n){
  vector<pint> ans;
  for(ll p=2;p<=sqrt(n);p++){
    if(n%p!=0) continue;
    ll cnt=0;
    while(n%p==0){
      n/=p;
      cnt++;
    }
    ans.push_back(make_pair(p,cnt));
  }
  if(n!=1) ans.push_back(make_pair(n,1));
  return ans;
}

/*↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓*/
int main() {
  IOS;
  ll n;
  cin>>n;
  vector<ll> v(n);
  rep(i,n) cin>>v[i];
  if(v[0]!=0){
    cout<<0<<endl;
    return 0;
  }
  sort(ALL(v));
  ll stock=v[n-1];
  ll ans=1;
  map<ll,ll> mp;
  rep(i,n) mp[v[i]]++;
  if(mp[0]>=2){
    cout<<0<<endl;
    return 0;
  }
  ll tmp=mp[1];
  for(ll i=2;i<=stock;i++){
    ans*=(ll)pow(tmp,mp[i])%MOD;
    tmp=mp[i];
    if(ans==0){
      cout<<0<<endl;
      return 0;
    }
  }
  
  cout<<ans%MOD<<endl;
}