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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)


import java.util.Scanner;

public class Main {
    private static long MOD = 998244353;

    public static void main(final String[] args) {
        final Scanner scanner = new Scanner(System.in);
        final int n = scanner.nextInt();
        final long[] array = new long[n + 1];
        for (int i = 0; i < n; i++) {
            final int tmp = scanner.nextInt();
            if (i == 0 && tmp != 0) {
                System.out.println(0);
                return;
            }
            array[tmp]++;
        }

        int last = n;
        for (; last >= 0; last--) {
            if (array[last] != 0) {
                break;
            }
        }

        for (int i = 0; i <= last; i++) {
            if (array[i] == 0) {
                System.out.println(0);
                return;
            }
        }

        long answer = 1;
        long pre = 1;
        for (int i = 1; i <= last; i++) {
            answer *= originalPow(pre, array[i]);
            answer %= MOD;
            pre = array[i];
        }

        System.out.println(answer);
    }

    private static long originalPow(final long a, final long b) {
        long result = 1;
        for (long i = 0; i < b; i++) {
            result *= a % MOD;
            result %= MOD;
        }

        return result;
    }
}