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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

#!/usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
from bisect import bisect_left, bisect_right
import sys, random, itertools, math
sys.setrecursionlimit(10**5)
input = sys.stdin.readline
sqrt = math.sqrt
def LI(): return list(map(int, input().split()))
def LF(): return list(map(float, input().split()))
def LI_(): return list(map(lambda x: int(x)-1, input().split()))
def II(): return int(input())
def IF(): return float(input())
def LS(): return list(map(list, input().split()))
def S(): return list(input().rstrip())
def IR(n): return [II() for _ in range(n)]
def LIR(n): return [LI() for _ in range(n)]
def FR(n): return [IF() for _ in range(n)]
def LFR(n): return [LI() for _ in range(n)]
def LIR_(n): return [LI_() for _ in range(n)]
def SR(n): return [S() for _ in range(n)]
def LSR(n): return [LS() for _ in range(n)]
mod = 998244353
inf = float('INF')

#A
def A():
    n = II()
    n -= 1
    print(n // 2)
    return

#B
def combination_mod(n, k, mod=mod):
    """ power_funcを用いて(nCk) mod p を求める """ 
    """ nCk = n!/((n-k)!k!)を使用 """

    from math import factorial
    if n < 0 or k < 0 or n < k: return 0
    if n == 0 or k == 0: return 1
    a = factorial(n) % mod
    b = factorial(k) % mod
    c = factorial(n - k) % mod
    return (a * pow(b, mod - 2, mod) * pow(c, mod - 2, mod)) % mod
    

def B():
    n = II()
    d = LI()
    dic = defaultdict(int)
    if d[0] != 0:
        print(0)
        return
    for di in d[1:]:
        dic[di] += 1
    ans = 1
    di = list(dic.items())
    di.sort()
    if di[0][0] != 1:
        print(0)
        return
    for i in range(len(di) - 1):
        if di[i][0] != di[i + 1][0] - 1:
            print(0)
            return
    be = 1
    for _, b in di:
        ans *= pow(be, b, mod)
        be = b
    print(ans)
    return

#C
def C():
    return

#D
def D():
    return

#E
def E():
    return

#F
def F():
    return

#G
def G():
    return

#H
def H():
    return

#Solve
if __name__ == '__main__':
    B()