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;
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define ALL(A) A.begin(),A.end()
#define RALL(A) A.rbegin(),A.rend()
typedef long long LL;
typedef pair<int,int> P;
const LL mod=998244353;
const LL LINF=1LL<<60;
const int INF=1<<30;
int dx[]={1,0,-1,0};
int dy[]={0,1,0,-1};


LL fpow(LL a, LL n){
    LL x = 1;
    while(n > 0){
        if(n&1){
            x=x*a%mod;
        }
        a=a*a%mod;
        n >>= 1;
    }
    return x;
}

int main(){
    int n;cin >> n;
    vector<int> d(n);
    vector<LL> k(n+1,0);
    for (int i = 0; i < n; i++) {
        cin >> d[i];
        k[d[i]]++;
    }
    if(d[0]!=0||k[0]!=1){
        cout << 0 << endl;
        return 0;
    }
    LL ans = 1;
    bool f = false;
    for (int i = 1; i < n; i++) {
        if(f&&k[i]){
            cout<<0<< endl;
            return 0;
        }
        if(k[i]==0){
            f = true;
            continue;
        }
        ans = (ans * fpow(k[i-1],k[i]))%mod;
    }
    cout << ans << endl;
    return 0;
}