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 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#define REP(i,m,n) for(int i=(m); i<(int)(n); i++)
#define RREP(i,m,n) for(int i=(int)(n-1); i>=m; i--)
#define rep(i,n) REP(i,0,n)
#define rrep(i,n) RREP(i,0,n)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define fi first
#define se second
#define debug(...) {cerr<<"[L"<<__LINE__<<"] "; _debug(__VA_ARGS__);}

template<typename T>
string join(const vector<T>&v, string del=", "){ stringstream s;
    for(auto x : v) s << del << x; return s.str().substr(del.size());
}
template<typename T>
ostream& operator<<(ostream& o, const vector<T>&v){
    if(v.size()) o << "[" << join(v) << "]"; return o;
}
template<typename T>
ostream& operator<<(ostream& o, const vector<vector<T> >&vv){
    int l = vv.size();
    if(l){ o<<endl; rep(i,l) o << (i==0 ? "[ " : ",\n  " ) << vv[i] << (i==l-1 ? " ]" : ""); }
    return o;
}
template<typename T1, typename T2>
ostream& operator<<(ostream& o, const pair<T1, T2>& p){
    return o << "(" << p.first << ", " << p.second << ")";
}
inline void _debug(){cerr<<endl;}
template<class First, class... Rest>
void _debug(const First& first, const Rest&... rest){cerr<<first<<" ";_debug(rest...);}

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;

const double PI = (1*acos(0.0));
const double EPS = 1e-9;
const int INF = 0x3f3f3f3f;
const ll INFL = 0x3f3f3f3f3f3f3f3fLL;
const ll mod = 998244353;

inline void finput(string filename) {
    freopen(filename.c_str(), "r", stdin);
}

struct mint {
    long long x;
    mint(long long x=0) : x((x%mod+mod)%mod) {}
    mint operator-() const {
      return mint(-x);
    }
    mint& operator+=(const mint& a) {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator-=(const mint& a) {
        if ((x += mod-a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator*=(const  mint& a) {
        (x *= a.x) %= mod;
        return *this;
    }
    mint operator+(const mint& a) const {
        mint res(*this);
        return res+=a;
    }
    mint operator-(const mint& a) const {
        mint res(*this);
        return res-=a;
    }
    mint operator*(const mint& a) const {
        mint res(*this);
        return res*=a;
    }
    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }
    mint inv() const {
        return pow(mod-2);
    }
    mint& operator/=(const mint& a) {
        return (*this) *= a.inv();
    }
    mint operator/(const mint& a) const {
        mint res(*this);
        return res/=a;
    }
    friend ostream& operator<<(ostream& os, const mint& m){
        os << m.x; return os;
    }
};

int main(){
    ios_base::sync_with_stdio(0);
    // finput("./input");
    int n; cin >> n;
    int lim = -1;
    vector<mint> nums(n+1);
    rep(i,n){
        int d; cin >> d;
        nums[d] += 1;
        lim = max(lim, d);
    }
    mint ans = 1;
    rep(i,lim){
        ans *= nums[i].pow(nums[i+1].x);
    }
    cout << ans << endl;
    return 0;
}