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>
#include <cstdio>
#include <stdio.h>
#include <algorithm>
#include <iomanip>
#include <cstdlib>
#include <vector>
#include <set>
#include <map>
#include <string>
#include <queue>
#include <bitset>
#include <cmath>
#include <ctime>
#define ff first
#define ss second
#define i1 fgvhdsjklfgh
#define j1 hvfdjsklfkgh
#define MAXN 100005
#define INF 1000000000000000071LL
#define ll long long
#define int long long

using namespace std;

const int mod = 1000000007;
int n, ans = 1, i, a[MAXN], cnt[MAXN];

main(){
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    //freopen("input.txt", "r", stdin);
    //freopen("terminator.in", "r", stdin);
    //freopen("terminator.out", "w", stdout);

    cin >> n;

    for(i = 0; i < n; i++){
        cin >> a[i];
        cnt[a[i]]++;
    }

    sort(a, a + n);

    if(a[0] != 0 || cnt[0] > 1){
        cout << 0;
        return 0;
    }

    for(i = 1; i < n; i++){
        //cout << a[i] << " " << cnt[a[i] - 1] << endl;

        ans = (ans * cnt[a[i] - 1]) % 998244353;
    }

    cout << ans;

    return 0;
}