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: Wrong Answer)

#include <bits/stdc++.h>
#define REP(i, a, n) for (long long i = (a); i < (long long)(n); ++i)
#define REPC(i, a, n) for (long long i = (a); i <= (long long)(n); ++i)
#define ALL(t) t.begin(), t.end()
#define RALL(t) t.rbegin(), t.rend()
#define MATINIT(type, row, col, init) \
  vector<vector<type>>(row, vector<type>(col, init));
#define Yes(cond) cout << (cond ? "Yes" : "No") << endl;
#define YES(cond) cout << (cond ? "YES" : "NO") << endl;
using namespace std;
using LL = long long;
using ULL = unsigned long long;
template <class T>
using VEC = std::vector<T>;
template <class T>
using MAT = std::vector<std::vector<T>>;
void DUMP() { cerr << endl; }
template <class Head, class... Tail>
void DUMP(Head &&head, Tail &&... tail) {
  cerr << head << ", ";
  DUMP(std::move(tail)...);
}
template <typename T>
ostream &operator<<(ostream &os, vector<T> &vec) {
  os << "{";
  for (auto v : vec) os << v << ",";
  os << "}";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, set<T> &s) {
  os << "{";
  for (auto p : s) os << p << ",";
  os << "}";
  return os;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, map<T1, T2> &m) {
  os << "{";
  for (auto p : m) os << p << ",";
  os << "}";
  return os;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, pair<T1, T2> p) {
  os << "[" << p.first << " " << p.second << "]";
  return os;
}
/* calculate n^p mod M. O(log(p)) */ long long modpow(long long n, long long p,
                                                      long long M) {
  long long ans = 1;
  while (p > 0) {
    if (p & 1) ans = (ans * n) % M;
    n = (n * n) % M;
    p >>= 1;
  }
  return ans;
}
LL M = 998244353;

int main() {
  int N;
  cin >> N;
  VEC<int> D(N);
  VEC<int> C(N, 0);
  REP(i, 0, N) {
    cin >> D[i];
    C[D[i]]++;
  }
  int tmpcnt = 0;
  REP(i, 0, N) {
    if (C[i] == 0) break;
    tmpcnt += C[i];
  }
  if (tmpcnt != N) {
    cout << 0 << endl;
    return 0;
  }
  // DUMP(C);
  LL ans = 1;
  LL child_num = 1;
  REP(i, 1, N) {
    ans = (ans * modpow(child_num, C[i], M)) % M;
    child_num = C[i];
  }
  cout << ans << endl;

  return 0;
}