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 Rust (1.15.1) (judge verdict: Runtime Error)

#![allow(unused_imports)]
#![allow(unused_variables)]
#![allow(non_snake_case)]
#![allow(dead_code)]
#![allow(deprecated)]

use std::cell::{Cell, Ref, RefCell, RefMut};
use std::cmp::{max, min, Ordering};
use std::collections::*;
use std::fmt::{Debug, Formatter, Write as FmtWrite};
use std::io::{stderr, stdin, BufRead, Write};
use std::mem::{replace, swap};
use std::ops::*;
use std::rc::Rc;
use std::usize;

const MOD_10_9_7: u64 = 1_000_000_007;
const INF: i64 = 1_000_000_000_000;
const MIN_INF: i64 = -1_000_000_000_000;

/// FYI: https://github.com/vain0x/scan-bench
#[allow(unused_macros)]
macro_rules! read {
    ([$t:ty] ; $n:expr) =>
        ((0..$n).map(|_| read!([$t])).collect::<Vec<_>>());
    ($($t:ty),+ ; $n:expr) =>
        ((0..$n).map(|_| read!($($t),+)).collect::<Vec<_>>());
    ([$t:ty]) =>
        (rl().split_whitespace().map(|w| w.parse().unwrap()).collect::<Vec<$t>>());
    ($t:ty) =>
        (rl().parse::<$t>().unwrap());
    ($($t:ty),*) => {{
        let buf = rl();
        let mut w = buf.split_whitespace();
        ($(w.next().unwrap().parse::<$t>().unwrap()),*)
    }};
}

fn rl() -> String {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    buf.trim_right().to_owned()
}

#[allow(unused_macros)]
macro_rules! debug {
    ($($format:tt)*) => (write!(std::io::stderr(), $($format)*).unwrap());
}

#[allow(unused_macros)]
macro_rules! debugln {
    ($($format:tt)*) => (writeln!(std::io::stderr(), $($format)*).unwrap());
}

fn a() {
    let n = read!(usize);
    if n <= 2 {
        println!("0");
        return;
    }
    println!("{}", ((n - 1) as f64 / 2 as f64).floor() as u64);
}

/// 累乗のmod
pub fn mod_pow(x: u64, n: u64, m: u64) -> u64 {
    let mut res = 1;
    let mut x = x % m;
    let mut n = n;
    while n > 0 {
        if n & 1 == 1 {
            res = (res * x) % m;
        }
        x = (x * x) % m;
        n >>= 1;
    }
    res
}

fn b() {
    let n = read!(usize);
    let d: Vec<usize> = read![[usize]];

    if d[0] > 0 {
        println!("0");
        return;
    }

    if n == 1 {
        println!("1");
        return;
    }

    let mut count = vec![0; n - 1];
    for i in 0..d.len() {
        if i > 0 && d[i] == 0 {
            println!("0");
            return;
        }
        count[d[i]] += 1;
    }

    let mut ans = 1u64;
    let mo = 998244353u64;
    for i in 1..(n - 1) {
        if i == 1 {
            // ans += count[i] % mo;
            continue;
        } else {
            let tmp = mod_pow(count[i - 1], count[i], mo);
            ans = ans % mo;
            ans *= tmp;
        }
    }

    println!("{}", ans);
}

fn main() {
    b();
}