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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

// ----------- define --------------
#define int long long
#define vi vector<int>
#define ii pair<int,int>
#define fi first
#define sc second
#define stoi stoll
#define popcnt __builtin_popcount
#define getBit(x, k) ((x >> k) & 1)
#define all(x) (x).begin(),(x).end()
// ---------------------------------

const int mod = 998244353;

#define pow abcdef_pow
int pow(int a, int b) {
  if (b == 0) return 1;
  int t = pow(a, b / 2);
  t = (t * t) % mod;
  if (b % 2) return (t * a) % mod;
  return t;
}

void Main() {
  int n; cin >> n;
  vector<int> a(n);
  vector<int> m(n);
  int mx = 0;
  for (int i = 0; i < n; i++) {
    int t; cin >> t;
    m[t]++;
    mx = max(mx, t);
  }
  int ans = (m[0] == 1);
  for (int i = 1; i <= mx; i++) {
    ans *= pow(m[i - 1], m[i]);
    ans %= mod;
  }
  cout << ans << '\n';
}

signed main() {
#ifdef _DEBUG
  // freopen("in" , "r", stdin );
  cerr << "- ---- -- ----- <3\n";
#endif
  cin.tie(0)->sync_with_stdio(0);
  int T = 1;
  // cin >> T;
  while (T--) Main();
}