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>) {
    val mod = 998244353L
    val n = readLine()!!.toInt()
    val dList = readLine()!!.split(' ').map(String::toLong).sorted()

    if (n == 1 && dList[0] == 0L) {
        println(1)
        return
    } else if (n == 1) {
        println(0)
        return
    }

    val isOk = (0 until n - 1).all { dList[it] == dList[it + 1] || dList[it] + 1 == dList[it + 1] }
            && dList[0] == 0L
            && dList[1] == 1L
    if (isOk.not()) {
        println(0)
        return
    }

    var ans = 1L
    val dG = dList.groupBy { it }
    var before = dG[1]!!.size.toLong()
    (2..dList.last()).forEach {
        val next = dG[it]!!.size.toLong()
        for (i in 1..next) {
            ans *= before
            ans %= mod
        }
        before = next
    }
    println(ans)
}