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)

#include <iostream>
using namespace std;


long long int modexpo(long long int a, long long int b, long long int m) 
{ 
    
    if (a == 0) 
        return 0; 
    if (b == 0) 
        return 1; 
  
   
    if (b % 2 == 0) { 
        long long int  ans; 
        ans = modexpo(a, b / 2, m); 
        ans = (ans * ans) % m; 
        ans=(ans+m)%m;
        return ans;
    } 
  
   
    else { 
        long long int  ans=a; 
        ans = ((ans%m) * (modexpo(a, b - 1, m) % m)) % m; 
        ans=(ans+m)%m;
        return ans;
    } 
  
    
} 

int main() {
	long long int n;
	cin>>n;
    long long int a[n];
    for(int i=0;i<n;i++){
        cin>>a[i];
    }
    long long int m=998244353;
    long long int b[n];
    for(int i=0;i<n;i++){
        b[i]=0;
    }
    for(int i=0;i<n;i++){
        b[a[i]]++;
    }
    if(b[0]!=1){
        cout<<0<<endl;
        return 0;
    }
    long long int ans=1,limit=n;
    
    for(int i=0;i<n;i++){
        if(b[i]==0){
            limit=i;
            break;
        }
    }
    for(int i=limit+1;i<n;i++){
        if(b[i]!=0){
            cout<<0<<endl;
            return 0;
        }
    }
  //   cout<<limit<<endl;
    for(int i=2;i<limit;i++){
        long long int temp=modexpo(b[i-1],b[i],m);
        ans=(ans*temp)%m;
    }
    cout<<ans<<endl;
}