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 <vector>
#include <algorithm>

using namespace std;

int MAX = 998244353;

long long int power(long long int value, long long int times) {
    if (times == 1)
        return value % MAX;
    long long int half = power(value, times/2);
    if (times&1)
        return (((half*half) % MAX) * half ) % MAX; 
    return (half*half) % MAX;
}

int main(){
   
    int N;
    cin >> N;

    vector<int> vertex(N);

    for (int i = 0; i < N; i++)
        cin >> vertex[i];


    sort(vertex.begin(), vertex.end());

    int previous_count = 0;

    bool fail = false;
    long long int ways = 1;

    int idx;

    for (idx = 0; idx < N; idx++)
        if (vertex[idx] != 0)
            break;
        else
            previous_count++;

    if (previous_count != 1)
        fail = true;
    else {
        int value = vertex[1];
        int count = 0;
        for (idx = 1; idx < N; idx++) {
            if (vertex[idx] != value) {
               
                if (vertex[idx] != value+1) {
                    fail = true;
                    break;
                }

                (ways *= power(previous_count, count)) %= MAX;

                previous_count = count;
                count = 1;
                value = vertex[idx];
            } else
                count++;
        }

        (ways *= power(previous_count, count)) %= MAX;  
    }

    if (fail)
        cout << 0;
    else
        cout << ways;
    cout << "\n";

    return 0;
}