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 <bits/stdc++.h>

using namespace std;

#define INF 998244353
#define int long long
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define rep(i, n) FOR(i, 0, n)
int gcd(int a, int b)
{
    return b ? gcd(b, a % b) : a;
}
int lcm(int a, int b) { return a / gcd(a, b) * b; }
signed main()
{
    int N;
    cin >> N;
    vector<int> D;
    vector<int> A;
    int ans = 1;
    rep(i, N)
    {
        int a;
        cin >> a;
        D.push_back(a);
    }
    if (D[0] != 0)
    {
        cout << "0" << endl;
        return 0;
    }
    sort(D.begin(), D.end());
    if (D[1] == 0)
    {
        cout << "0" << endl;
        return 0;
    }
    int j = 0;
    int count = 0;
    for (int i = 0; i < N; i++)
    {
        count = 0;
        for (; j < N; j++)
        {
            if (D[j] == i)
            {
                count++;
            }
            else
            {
                if (count == 0)
                {
                    cout << "0" << endl;
                    return 0;
                }
                A.push_back(count);
                break;
            }
            if (j == N - 1)
            {
                A.push_back(count);
            }
        }
    }
    for (int i = 1; i < A.size(); i++)
    {
        ans = ((ans % INF) * ((int)pow(A[i - 1], A[i]) % INF)) % INF;
    }
    cout << ans << endl;
}