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)

//  atcodder.xcode
//  main.cpp
//  Created by conan1024hao on 2019/11/09.
//  Copyright © 2019 conan1024hao. All rights reserved.
//  専用ライブラリ
#pragma GCC optimize ("O3")
#include <iostream>
#include <iomanip>
#include <istream>
#include <ostream>
#include <sstream>
#include <iterator>
#include <vector>
#include <algorithm>
#include <queue>
#include <deque>
#include <list>
#include <stack>
#include <map>
#include <set>
#include <utility>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <ctime>
#include <cctype>
#include <cstdlib>
using namespace std;
typedef long long int ll;
typedef pair<ll,ll>P;
#define INF 10e16
#define MOD 1000000007
#define rep(i, a, n) for (ll i = a; i < (ll)(n); i++)
#define fi first
#define se second
#define mmax(x,y)(x>y?x:y)
#define mmin(x,y)(x<y?x:y)
#define chmax(x,y) x=mmax(x,y)
#define chmin(x,y) x=mmin(x,y)
#define all(x) (x).begin(),(x).end()
#define PI acos(-1.0)
long long GCD(long long a, long long b) { return b ? GCD(b, a%b) : a; }
long long LCM(long long a, long long b)  {return a/GCD(a,b)*b;}
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
ll cmp(pair<ll,string>a,pair<ll,string> b){
        if(a.fi!=b.fi)
        return a.fi<b.fi;
        else
        return a.se<b.se;
}
//---------------------------------------------------------------------------

//---------------------------------------------------------------------------
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    //-------------------------------
    int n;
    cin>>n;
    vector<int>d(n);
    for(int i=0;i<n;i++){
        cin>>d[i];
    }
    int cnt[100001]={0};
    bool abc=false;
    if(d[0]>0)
        abc=true;
    ll ans=1;
    for(int i=1;i<n;i++){
        cnt[d[i]]++;
        if(d[i]==0)
            abc=true;
        else
            ans*=d[i];
        ans%=MOD;
    }
    ans%=998244353;
    if(abc==true)
        cout<<0<<endl;
    else
    cout<<ans<<endl;
    //-------------------------------
    return 0;
}
//---------------------------------------------------------------------------