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 <cassert>
#include <cstdio>
#include <cmath>
#include <iostream>
#include <iomanip>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <numeric>
#include <algorithm>

using namespace std;
using lint = long long;
constexpr int MOD = 998244353, INF = 1010101010;
constexpr lint LINF = 1LL << 60;

template <class T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
	for (const auto &e : vec) os << e << (&e == &vec.back() ? "\n" : " ");
	return os;
}

#ifdef _DEBUG
template <class T>
void dump(const char* str, T &&h) { cerr << str << " = " << h << "\n"; };
template <class Head, class... Tail>
void dump(const char* str, Head &&h, Tail &&... t) {
	while (*str != ',') cerr << *str++; cerr << " = " << h << "\n";
	dump(str + (*(str + 1) == ' ' ? 2 : 1), t...);
}
#define DMP(...) dump(#__VA_ARGS__, __VA_ARGS__)
#else 
#define DMP(...) ((void)0)
#endif

template<int Modulo = MOD>
struct Mint {

	lint val;
	constexpr Mint(lint v = 0) noexcept : val(v % Modulo) { if (val < 0) val += Modulo; }

	constexpr Mint& operator += (const Mint &r) noexcept {
		val += r.val;
		if (val >= Modulo) val -= Modulo;
		return *this;
	}
	constexpr Mint& operator -= (const Mint &r) noexcept {
		val -= r.val;
		if (val < 0) val += Modulo;
		return *this;
	}
	constexpr Mint& operator *= (const Mint &r) noexcept {
		val = val * r.val % Modulo;
		return *this;
	}
	constexpr Mint& operator /= (const Mint &r) noexcept {
		lint a = r.val, b = Modulo, u = 1, v = 0;
		while (b) {
			lint t = a / b;
			a -= t * b; swap(a, b);
			u -= t * v; swap(u, v);
		}
		val = val * u % Modulo;
		if (val < 0) val += Modulo;
		return *this;
	}

	constexpr Mint operator + (const Mint &r) const noexcept { return Mint(*this) += r; }
	constexpr Mint operator - (const Mint &r) const noexcept { return Mint(*this) -= r; }
	constexpr Mint operator * (const Mint &r) const noexcept { return Mint(*this) *= r; }
	constexpr Mint operator / (const Mint &r) const noexcept { return Mint(*this) /= r; }

	constexpr int getmod() { return Modulo; }
	constexpr Mint operator - () const noexcept { return val ? Modulo - val : 0; }

	constexpr bool operator == (const Mint &r) const noexcept { return val == r.val; }
	constexpr bool operator != (const Mint &r) const noexcept { return val != r.val; }

	friend ostream &operator << (ostream &os, const Mint<Modulo> &x) noexcept { return os << x.val; }
	friend istream &operator >> (istream &is, Mint<Modulo> &x) noexcept {
		lint tmp; is >> tmp;
		x = Mint<Modulo>(tmp);
		return is;
	}
	friend constexpr Mint<Modulo> modpow(const Mint<Modulo> &a, lint n) noexcept {
		Mint res(1), tmp = a;
		while (n > 0) {
			if (n & 1) res *= tmp;
			tmp *= tmp;
			n >>= 1;
		}
		return res;
	}
};

using mint = Mint<>;

lint modpow(lint a, lint n, lint mod = MOD) {
	lint res = 1;
	while (n > 0) {
		if (n & 1) res = res * a % mod;
		a = a * a % mod;
		n >>= 1;
	}
	return res;
}

int main() {

	cin.tie(nullptr);
	ios::sync_with_stdio(false);

	int N;
	cin >> N;

	vector<int> D(N);
	for (int i = 0; i < N; i++) cin >> D[i];

	vector<int> dist(N);
	for (int i = 0; i < N; i++) dist[D[i]]++;
	
	mint ans = 1;
	if (D[0] != 0 || dist[0] != 1) ans = 0;

	for (int i = 1; i < N; i++) {
		if (dist[i]) ans *= modpow(dist[i-1], dist[i]);
		else {
			while (i < N) {
				if (dist[i]) ans = 0;
				i++;
			}
		}
	}

	cout << ans << "\n";

	return 0;
}