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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define rep(i,n) for(ll i=0;i<n;i++)
#define P pair<ll,ll>
#define Graph vector<vector<ll>>
#define INF (1ll<<60)
#define mod 998244353
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;}

ll modpow(ll x,ll n){
    ll ret=1;
    while(n>0){
        if(n&1) ret=ret*x%mod;  // n の最下位bitが 1 ならば x^(2^i) をかける
        x=x*x%mod;
        n>>=1;  // n を1bit 左にずらす
    }
    return ret;
}

int main(){
    ll n;cin>>n;
    vector<ll> d(n);
    rep(i,n) cin>>d[i];
    vector<ll> hm(100010,0);//距離ごとに個数を調べる
    
    sort(d.begin(),d.end());
    if(d[0]!=0){
        cout<<0<<endl;
        return 0;
    }
    rep(i,n){
        if(i!=0 && d[i]==0){
            cout<<0<<endl;
            return 0;
        }
    }
    rep(i,n-1){
        if(d[i+1]-d[i]>1){
            cout<<0<<endl;
            return 0;
        }
    }


    rep(i,n){//count
        hm[d[i]]++;
    }

    ll ans=1;
    rep(i,100001){
        if(hm[i+1]==0) break;
    
        ans*=modpow(hm[i],hm[i+1]);
        ans%=mod;
    }

    cout<<ans<<endl;
    return 0;
}