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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

fun mPow(n: Long, ex: Long, m: Long): Long {
    var x = n
    var remain = ex
    var result = 1L
    while (remain > 0) {
        if (remain % 2L == 1L) {
            result = (result * x) % m
        }
        x = (x * x) % m
        remain /= 2
    }
    return result
}

fun main(args: Array<String>) {
    val n = readLine()?.toInt() ?: return
    val d = readLine()?.split(" ")?.map(String::toInt) ?: return
    val m = 998244353L
    if (d.first() != 0) {
        println("0")
        return
    }
    var count = 1L
    var prev = 0L
    d.groupBy { it }.entries.sortedBy { it.key }.forEach { v ->
        if (v.key == 0 && v.value.size > 1) {
            println("0")
            return
        } else if (v.key > 1) {
            count = (count * mPow(prev, v.value.size.toLong(), m)) % m
        }
        prev = v.value.size.toLong()
    }
    println(count.toString())
}