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: Accepted)

#include <bits/stdc++.h>
#ifdef LOCAL_DEBUG
#define DEBUG 1
#define CERR if(DEBUG) cerr
#define MACRO(_1,_2,_3,_4,_5,_6,_7,_8,_9,_10,NAME,...) NAME
#define pr(...) CERR<<MACRO(__VA_ARGS__,pr10,pr9,pr8,pr7,pr6,pr5,pr4,pr3,pr2,pr1)(__VA_ARGS__)<<endl
#define pr1(a) (#a)<<"="<<(a)<<" "
#define pr2(a,b) pr1(a)<<pr1(b)
#define pr3(a,b,c) pr1(a)<<pr2(b,c)
#define pr4(a,b,c,d) pr1(a)<<pr3(b,c,d)
#define pr5(a,b,c,d,e) pr1(a)<<pr4(b,c,d,e)
#define pr6(a,b,c,d,e,f) pr1(a)<<pr5(b,c,d,e,f)
#define pr7(a,b,c,d,e,f,g) pr1(a)<<pr6(b,c,d,e,f,g)
#define pr8(a,b,c,d,e,f,g,h) pr1(a)<<pr7(b,c,d,e,f,g,h)
#define pr9(a,b,c,d,e,f,g,h,i) pr1(a)<<pr8(b,c,d,e,f,g,h,i)
#define pr10(a,b,c,d,e,f,g,h,i,j) pr1(a)<<pr9(b,c,d,e,f,g,h,i,j)
#define prArr(a){CERR<<(#a)<<"={";int i=0;for(auto t:(a))CERR<<(i++?", ":"")<<t;CERR<<"}"<<endl;}
#else
#define DEBUG 0
#define pr(...)
#define prArr(a)
#endif
using namespace std;
using Int = long long;
using _int = int;
using ll = long long;
using Double = long double;
const Int INF = (1LL<<60)+1e9; // ~ 1.15 * 1e18
const Int mod = 998244353;
const Double EPS = 1e-8;
const Double PI = 6.0 * asin((Double)0.5);
using P = pair<Int,Int>;
template<class T> T Max(T &a,T b){return a=max(a,b);}
template<class T> T Min(T &a,T b){return a=min(a,b);}
template<class T1, class T2> ostream& operator<<(ostream& o,pair<T1,T2> p){return o<<"("<<p.first<<","<<p.second<<")";}
template<class T1, class T2, class T3> ostream& operator<<(ostream& o,tuple<T1,T2,T3> t){return o<<"("<<get<0>(t)<<","<<get<1>(t)<<","<<get<2>(t)<<")";}
template<class T1, class T2> istream& operator>>(istream& i,pair<T1,T2> &p){return i>>p.first>>p.second;}
template<class T> ostream& operator<<(ostream& o,vector<T> a){Int i=0;for(T t:a)o<<(i++?" ":"")<<t;return o;}
template<class T> istream& operator>>(istream& i,vector<T> &a){for(T &t:a)i>>t;return i;}
//INSERT ABOVE HERE


signed main(){
  srand((unsigned)time(NULL));
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(12);

  Int N;
  cin>>N;
  vector<Int> D(N);
  cin>>D;
  Int ans = 1;
  if(D[0] != 0) ans = 0;

  sort(D.begin(), D.end());
  map<Int, Int> cnt;
  cnt[-1] = 1;

  for(Int i=0;i<N;i++){
    Int d = D[i];
    if(i != 0 && d == 0) ans = 0;
    ans *= cnt[d-1];
    ans %= mod;
    cnt[d]++;
    if(ans == 0) break;
  }

  cout<<ans<<endl;



  return 0;
}