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 main(args: Array<String>) {
    nikkei20192quala_b()
}

private fun mod(x: Long) = x % (998244353).toLong()

fun nikkei20192quala_b() {
    val n = readLine()!!.toInt()
    val dList = readLine()!!.split(" ").map { it.toInt() }

    val eachCount = dList.groupBy { it }.mapValues { it.value.size }

    if (eachCount[0] == null) return println(0)

    var answer = 1L
    var depth = 0
    while (eachCount[depth + 1] != null) {
        depth++
        answer *= mod(Math.pow(eachCount[depth - 1]?.toDouble() ?: 0.0, eachCount[depth]?.toDouble() ?: 0.0).toLong())
    }

    val last = eachCount.keys.max()!!

    if (depth < last) return println(0)

    println(answer)
}