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 <iostream>
#include <vector>
#include <algorithm>
#include <math.h>
using namespace std;

#define int long long

#if defined(__GNUC__)
#define __int32 __int32_t
#endif // __GNUC__

#define itn __int32
#define REP(i, n) for(itn (i) = 0; (i) < (n); (i)++)
#define REP2(i, x, n) for(itn (i) = x; (i) < (n); (i)++)
#define ALL(x) (x).begin(), (x).end()
#define RALL(x) (x).rbegin(), (x).rend()
#define SORT(x) sort(ALL(x))
#define RSORT(x, T) sort(ALL(x), greater<(T)>())
#define REVERCE(x) reverce(ALL(x))

itn main() {
	ios::sync_with_stdio(false);

	itn n;

	cin >> n;

	if (n == 1) {
		cout << 0;
		return 0;
	}

	vector<int> d(n);

	REP(i, n) {
		cin >> d[i];
	}

	int cnt = 1;
	
	vector<itn> ren_set;

	SORT(d);

	itn ren = 1;

	while(cnt < n){
		
		while (cnt < n && d[cnt] == d[cnt - 1]) {
			ren++;
			cnt++;
		}

		if (d[cnt] - 1 != d[cnt - 1]) {
			cout << 0;
			return 0;
		}

		ren_set.push_back(ren);
		if (cnt == n - 1) {
			ren_set.push_back(1);
		}
		ren = 1;
		cnt++;
	}

	if (ren_set[0] != 1) {
		cout << 0;
		return 0;
	}

	if (ren_set.size() == 1) {
		cout << 0;
		return 0;
	}

	long double res = ren_set[0];

	REP2(i, 1, ren_set.size()) {
		res *= powl(ren_set[i - 1], ren_set[i]);
	}

	cout << fmod(res, (double)998244353);

	return 0;
}