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<iostream>
#include<string>
#include<iomanip>
#include<cmath>
#include<vector>
#include<algorithm>

using namespace std;

#define int long long
#define endl "\n"

const long long INF = (long long)1e18;
const long long MOD = 998244353; 

string yn(bool f){return f?"Yes":"No";}
string YN(bool f){return f?"YES":"NO";}



long long power(long long x, long long n){
	long long ans = 1;
	for(;n;n>>=1,x*=x,ans%=MOD,x%=MOD)
		if(n&1)ans*=x;
	return ans%MOD;
}



signed main(){
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	cout<<fixed<<setprecision(10);
	
	int N, zero = 0, maximum = 0;
	vector<int> D;
	vector<int> con;
	
	cin>>N;
	
	D.resize(N);
	con.resize(N);
	
	for(int i = 0; i < N; i++){
		cin>>D[i];
		// if(!D[i])zero++;
		con[D[i]]++;
		maximum = max(maximum, D[i]);
	}
	
	if(D[0]){
		cout<<0<<endl;
	} else if(con[0] >= 2){
		cout<<0<<endl;
	} else {
		int ans = 1;
		
		for(int i = 1; i <= maximum; i++){
			if(con[i] == 0) ans = 0;
			// else ans *= power(i, con[i]);
			else ans *= power(con[i-1], con[i]);
			
			ans %= MOD;
		}
		
		cout<<ans<<endl;
	}
	
	
	return 0;
}