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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {

    public static void main(String[] args) {
        // write your code here
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        long[] array = new long[n];

        for (int i = 0; i < n; i++) {
            array[scanner.nextInt()]++;
        }

        if (n < 2 || array[2] == 0) {
            System.out.println(0);
            return;
        }

        long ans = 1;

        for (int i = 2; i < n; i++) {
            if (array[i - 1] == 0 && array[i] > 0) {
                System.out.println(0);
                return;
            }
            long v = pow(array[i - 1], array[i], 998244353);
            ans *= v;
            ans %= 998244353;
        }

        System.out.println(ans);
    }

    //return a^b mod M O(logB)
    private static long pow(long a, long b, int M) {
        long ret = 1;
        long tmp = a;
        while (b > 0) {
            if ((b & 1) == 1) ret = (ret * tmp) % M;
            tmp = (tmp * tmp) % M;
            b = b >> 1;
        }
        return ret;
    }
}