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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
class Program
{
	static void Main(string[] args)
	{
		int N = int.Parse(Console.ReadLine());
		long mod = 998244353;
		long[] B = new long[N+1];
 		long[] A = new long[N+1];
		string[] str = Console.ReadLine().Split();
		for(var i=0;i<N;i++){
			int a2 = int.Parse(str[i]);
			B[i] = a2;
			A[a2]++;
		}
		long ans = 0;
		long a1 = 0;
      	bool flag = true;
		if(B[0]==0){
			ans = 1;
			a1 = 1;
			for(var i=1;i<=N;i++){
				if(A[i] > 0){
                  	if(flag == false){
                      	ans = 0;
                    }
					ans = ans * Mpow(a1,A[i],mod) % mod;
					a1 = A[i];
                } else {
                  	flag = false;
                }
				//Console.WriteLine(A[i]+" "+a1+" "+ans);
            }
		}
		Console.WriteLine(ans);
	}
	
	public static long Mpow(long x,long n,long m)
	{
		long ans = 1;
		while(n > 0){
			if(n % 2 == 1){
				ans = ans * x % m;
				n -= 1;
			}
			x = x * x % m;
			n /= 2;
		}
		return ans;
	}
}