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>
#define pb push_back
#define all(v) (v).begin() , (v).end()
#define popcnt(x) __builtin_popcount(x)
#define inf 0x3f3f3f3f
#define watch(x) cout << (#x) << " is " << (x) << endl
#define rand() (rand() << 15 | rand())
using namespace std;
typedef long long  ll;
#define EPS 1e-9
#define PI acos(-1.0)
const int MOD = 998244353;

int mul(int a , int b , int m = MOD){
   return (1LL * (a % m) * (b % m)) % m;
}

int add(int a , int b , int m = MOD){
   return (1LL * (a % m) + (b % m)) % m;
}

int powMOD(int x , int n , int m = MOD){
   if(!n)return 1;
   int ret = 1;
   if(n & 1)ret = mul(x , powMOD(x , n - 1 , m) , m);
   else{
      int y = powMOD(x , n / 2 , m);
      ret = mul(y , y , m);
   }
   return ret;
}
int main() {
   ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
   int n;
   cin >> n;
   vector<int> a(n);
   for(int &x : a)
      cin >> x;

   sort(all(a));
   int curLevel = 0 , lst = 0 , ans = 1;
   for(int i = 0 ; i < n ; ++i){
      if(a[i] != curLevel){
         cout << "0\n";
         return 0;
      }
      int j = i;
      while (j < n && a[j] == curLevel){
         ++j;
      }
      int cnt = j - i;
      if(lst)ans = mul(ans , powMOD(lst , cnt , MOD));
      lst = cnt;
      i = j - 1;
      curLevel++;
   }

   cout << ans << "\n";
}