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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define fore(b,c) for(int val0=b;val0<c;val0++)
#define forr(k,c,s) for(int k=c;k<s;k++)
#define pb push_back
#define mmp make_pair
#define oset(T) tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>
using namespace __gnu_pbds;
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
typedef pair<int,int> ii;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<vi> vvi;
typedef long double ld;
typedef vector<vii> al;
typedef vector<ll> vl;
const int INF = 1e9;
const ll INFL = 1LL<<61;
const ll M = 998244353;
const int MN = 200200;
ll fr[MN];
ll mul(ll a, ll b) {return (a*b)%M;}
ll bp(ll b, ll p) {
	ll ac = 1;
	while(p) {
		if(p&1) {
			ac = mul(ac,b);
		}
		b = mul(b,b);
		p >>= 1;
	}
	return ac;
}
ll w[MN];
int main() {
	ios::sync_with_stdio(0);cout.precision(20);cout.tie(0);cin.tie(0);
	int n;
	cin >> n;
	for(int i=0;i<n;i++) {
		cin >> w[i];
		fr[w[i]]++;
	}
	ll tot = 1;
	if(fr[0] != 1) {
		tot = 0;
	}
	if(w[0] != 0) {
		tot = 0;
	}
	for(int i=1;i<n;i++) {
		tot = mul(tot,bp(fr[i-1],fr[i]));
	}
	cout << tot << '\n';
}