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 C (GCC 5.4.1) (judge verdict: Runtime Error)

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#define PI 3.141592653589793
#define MOD 1000000007

int main(){
	int n;
	scanf("%d", &n);
	int drm[n+1];
	for(int i=1; i<=n; i++){
		scanf("%d", &drm[i]);
	}

	if(drm[1] != 0){
		printf("0\n");
		exit(1);
	}
	else{
		/*
		for(int i=2; i<=n; i++){
			if(drm[i] == 0){
				printf("0\n");
				exit(2);
			}
		}
		*/

		int max=0;
		for(int i=2; i<=n; i++){
			if(max<drm[i]) max = drm[i];
		}

		double hen[max+1];
		for(int i=0; i<max+1; i++){
			hen[i] = 0;
		}
		for(int i=1; i<=n; i++){
			hen[drm[i]] = hen[drm[i]] + 1;
		}
		long int answer = 1;
		double before=1, now;
		for(int i=1; i<=max; i++){
			now = hen[i];
			answer =(long int) answer*pow(before, now);
			answer = answer%998244353;
			before = now;
		}

		/*
		int before=1, now;
		long int answer=1;
		for(int i=1; i<=max; i++){
			now = 0;
			for(int j=1; j<=n; j++){
				if(drm[j] == i) now++;
				//printf("i:%d, now:%d\n", i, now);
			}
			answer =(long int) answer*pow(before, now);
			answer = answer%998244353;
			before = now;
			//printf("%ld\n", answer);
		}
		*/

		printf("%ld\n", answer);
	}
	//printf("%d\n", answer);

	return 0;
}