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;

typedef vector<int> vi;
typedef long long int ll;
typedef pair<int,int> pi;
typedef unsigned long long int ull;

ll lsone(ll p){return(p & -p);}

int main(){
    std::ios::sync_with_stdio(false);
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n, d[100100];
    map<int, int> deg;
    ll ans, prev_num, cur_num, md;
    md = 998244353;
    vi dis;
    dis.clear();
    cin >> n;
    cin >> d[1];
    if (d[1] != 0){
        cout << 0;
        return 0;
    }
    for (int i=2;i<=n;i++){
        cin >> d[i];
        dis.push_back(d[i]);
    }
    sort(dis.begin(), dis.end());
    int i = 0;
    int prev = 0;
    prev_num = 1;
    ans = 1;
    while (i < dis.size()){
        int cur = dis[i];
        int ii = i;
        if (cur - prev != 1){
            cout << 0;
            return 0;
        }
        while (i < dis.size() && dis[i] == cur){
            i++;
        }
        cur_num = i - ii;
        for (int j=0;j<cur_num;j++){
            ans = (ans * prev_num) % md;
        }
        prev = cur;
        prev_num = cur_num;
    }
    cout << ans;
}