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<bits/stdc++.h>
using namespace std;
long long  a[100005];
const long long  MOD = 998244353;
long long  mult(long long  a , long long  b) {
    return (a%MOD*b%MOD)%MOD;
}
long long  power(long long  a , long long  b){
    //cout << a << " " << b << "\n";
    if(b == 0 ) return a;
    if(b == 1) return a;
    if(b%2){
        long long  k = power(a , b/2);
        return mult(mult(k , k ),a);
    }
    else{
        long long  k = power(a , b/2);
        return mult(k , k );
    }
}
long long  sd[1000005] = {0};
int  main(int argc , char* argv[]){
    long long  n;
    cin >> n;
    for(long long  i = 0 ; i < n ; i++) cin >> a[i];
    if(a[0] !=0 ){ cout << "0"; return 0;}
    long long  ans = 1;
    
    for(long long  i = 0 ; i < n ; i++)
        sd[a[i]]++;
    long long  f= 0 ;
   // long long  f= 0 ;
    /*for(long long  i = 0 ; i < n ;i++){
        if(sd[i] == 1 && f == 1) { cout << "0"; return 0;}
        else if(sd[i] == 0){ f=1;
        } 
    }*/
    for(long long i = 1 ; i < n;i++){
        if(a[i]==0) {
            f=1;break;
        }
    }
    if(f){cout << "0"; return 0;}
    f=0;

    for(long long  i = 0 ; i < n ;i++){
        if(sd[i] == 1 && f == 1) { cout << "0"; return 0;}
        else if(sd[i] == 0){ f=1;
        } 
    }
    for(long long  i = 1 ; i < n ; i++){
        if(sd[i] == 0) continue;

        ans = mult(ans , power(sd[i-1] , sd[i]));
    }
    cout << ans;
    return 0;
    
}