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;

typedef long long ll;
const ll INF = 1LL<<60;

ll gcd(ll a, ll b) {
  if(b == 0) return a;
  return gcd(b, a % b);
}

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

ll modpow(ll a, ll n, ll mod) {
  ll res = 1;
  while (n > 0) {
    if (n & 1) res = res * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return res;
}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  int n;
  cin >> n;

  vector<int> cnt(n);

  ll MOD = 998244353;
  ll ans;

  int d0;
  cin >> d0;

  int tmp;
  int len_max = 0;
  for(int i = 1; i < n; i++){
    cin >> tmp;
    cnt[tmp]++;
    chmax(len_max, tmp);
  }

  if(d0 > 0 && cnt[0] > 0){
    cout << 0 << endl;
  }else{
    ll ans = 1;

    for(int i = 2; i <= len_max; i++){
      for(int j = 0; j < cnt[i]; j++){
        ans *= cnt[i-1];
        ans = ans % MOD;
      }
    }

    cout << ans << endl;
  }

  return 0;
}