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++14 (GCC 5.4.1) (judge verdict: Accepted)

/**
 * @copyright (c) 2020 Daisuke Hashimoto
 */

#include <bits/stdc++.h>
using namespace std;
using ll = int64_t;
using Pair = pair<int64_t, int64_t>;

// std::cout << std::setprecision(20) << 1.1 << endl;

// const int64_t kMod = 1000000007LL;
const int64_t kMod = 998244353LL;
struct Mint {
  int64_t value;
  Mint(int64_t value = 0LL) : value((value % kMod + kMod) % kMod) {}
  Mint operator-() const {
    return Mint(-value);
  }
  Mint &operator+=(const Mint rhs) {
    if ((value += rhs.value) >= kMod) {
      value -= kMod;
    }
    return *this;
  }
  Mint &operator-=(const Mint rhs) {
    if ((value += kMod - rhs.value) >= kMod) {
      value -= kMod;
    }
    return *this;
  }
  Mint &operator*=(const Mint rhs) {
    (value *= rhs.value) %= kMod;
    return *this;
  }
  Mint operator+(const Mint rhs) const {
    return Mint(*this) += rhs;
  }
  Mint operator-(const Mint rhs) const {
    return Mint(*this) -= rhs;
  }
  Mint operator*(const Mint rhs) const {
    return Mint(*this) *= rhs;
  }
  Mint Pow(int64_t p) const {
    if (p == 0LL) {
      return 1LL;
    } else if (p < 0LL) {
      return Pow(-p).Pow(kMod - 2LL);
    }
    Mint temp = Pow(p >> 1LL);
    temp *= temp;
    if (p & 1LL) {
      temp *= (*this);
    }
    return temp;
  }
  Mint Inv() const {
    return Pow(kMod - 2LL);
  }
  Mint &operator/=(const Mint a) {
    return *this *= a.Inv();
  }
  Mint operator/(const Mint a) const {
    return Mint(*this) /= a;
  }
};
std::ostream &operator<<(std::ostream &os, const Mint &a) {
  os << a.value;
  return os;
}

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  int64_t N;
  cin >> N;
  vector<int64_t> H(N, 0);
  int64_t depth = 0;
  for (int64_t i = 0; i < N; ++i) {
    int64_t d;
    cin >> d;
    if (i == 0 && d != 0) {
      cout << 0 << endl;
      return 0;
    } else if (i > 0 && d == 0) {
      cout << 0 << endl;
      return 0;
    }
    ++H[d];
    depth = max(d, depth);
  }
  if (H[0] != 1) {
    cout << 0 << endl;
    return 0;
  }
  for (int64_t i = 0; i <= depth; ++i) {
    if (H[i] == 0) {
      cout << 0 << endl;
      return 0;
    }
  }
  Mint result(1);
  for (int64_t i = 1; i <= depth; ++i) {
    Mint temp = H[i - 1];
    result *= temp.Pow(H[i]);
  }
  cout << result << endl;
  return 0;
}