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>

using namespace std;

const int maxn = 1e5 + 5;
const long long mod = 998244353;

long long d[maxn], vf[maxn], fact[maxn];

long long x, y;

void invmod(long long a, long long n, long long &x, long long &y) {
	if (n == 0) {
		x = 1, y = 0;
		return;
	}
	long long x1, y1;
	invmod(n, a % n, x1, y1);
	x = y1;
	y = x1 - (a / n) * y1;
}

long long findinv(long long nr)  {
    invmod(nr, mod, x, y);
    x %= mod;
    while(x < 0)
        x += mod;
    return x;
}

long long put(long long a, long long n)  {
    long long ans = 1LL;
    while(n)  {
        if(n % 2)  {
            ans *= a;
            ans %= mod;
        }
        a *= a;
        a %= mod;
        n /= 2;
    }
    return ans;
}

int main()  {
	int n;
	cin >> n;
	for(int i = 1;i <= n;i++)
        cin >> d[i], vf[d[i]]++;
    if(d[1] != 0 or vf[0] > 1)  {
        cout << "0";
        return 0;
    }
    fact[1] = 1;
    fact[0] = 1;
    for(int i = 2;i <= n;i++)
        fact[i] = fact[i - 1] * i, fact[i] %= mod;
    long long ans = 1LL;
    int ram = n - 1;
    vf[0] = 1;
    for(int i = 1;i <= n;i++)  {
        ram -= vf[i];
        if(ram > 0 && vf[i] == 0)  {
            cout << "0";
            return 0;
        }
        ans *= put(vf[i - 1], vf[i]);
        ans %= mod;
    }
    cout << ans;
	return 0;
}