Problem:
Score : 600 points

Problem Statement

You are given an integer sequence of length n+1, a_1,a_2,...,a_{n+1}, which consists of the n integers 1,...,n.
It is known that each of the n integers 1,...,n appears at least once in this sequence.

For each integer k=1,...,n+1, find the number of the different subsequences (not necessarily contiguous) of the given sequence with length k, modulo 10^9+7.

Notes

If the contents of two subsequences are the same, they are not separately counted even if they originate from different positions in the original sequence.

A subsequence of a sequence a with length k is a sequence obtained by selecting k of the elements of a and arranging them without changing their relative order. For example, the sequences 1,3,5 and 1,2,3 are subsequences of 1,2,3,4,5, while 3,1,2 and 1,10,100 are not.

Constraints

1 \leq n \leq 10^5

1 \leq a_i \leq n

Each of the integers 1,...,n appears in the sequence.

n and a_i are integers.

Input

Input is given from Standard Input in the following format:

n
a_1 a_2 ... a_{n+1}

Output

Print n+1 lines.
The k-th line should contain the number of the different subsequences of the given sequence with length k, modulo 10^9+7.

Sample Input 1

3
1 2 1 3

Sample Output 1

3
5
4
1

There are three subsequences with length 1: 1 and 2 and 3.

There are five subsequences with length 2: 1,1 and 1,2 and 1,3 and 2,1 and 2,3.

There are four subsequences with length 3: 1,1,3 and 1,2,1 and 1,2,3 and 2,1,3.

There is one subsequence with length 4: 1,2,1,3.

Sample Input 2

1
1 1

Sample Output 2

1
1

There is one subsequence with length 1: 1.

There is one subsequence with length 2: 1,1.

Sample Input 3

32
29 19 7 10 26 32 27 4 11 20 2 8 16 23 5 14 6 12 17 22 18 30 28 24 15 1 25 3 13 21 19 31 9

Sample Output 3

32
525
5453
40919
237336
1107568
4272048
13884156
38567100
92561040
193536720
354817320
573166440
818809200
37158313
166803103
166803103
37158313
818809200
573166440
354817320
193536720
92561040
38567100
13884156
4272048
1107568
237336
40920
5456
528
33
1

Be sure to print the numbers modulo 10^9+7.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#pragma GCC optimize("Ofast","unroll-loops")
#pragma GCC push_options
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#include <xmmintrin.h>
#include <immintrin.h>
using namespace::std;
__attribute__((constructor))void init(){cin.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(15);}
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/priority_queue.hpp>
#include<ext/pb_ds/tag_and_trait.hpp>
template<typename T>using pbds=__gnu_pbds::tree<T,__gnu_pbds::null_type,less<T>,__gnu_pbds::rb_tree_tag,__gnu_pbds::tree_order_statistics_node_update>;
template<typename T>using pbds_map=__gnu_pbds::tree<T,T,less<T>,__gnu_pbds::rb_tree_tag,__gnu_pbds::tree_order_statistics_node_update>;
template<typename T,typename E>using hash_map=__gnu_pbds::gp_hash_table<T,E>;
template<typename T>using pqueue =__gnu_pbds::priority_queue<T, less<T>,__gnu_pbds::rc_binomial_heap_tag>;
typedef long long lint;
#define INF (1LL<<60)
#define IINF (1<<30)
#define EPS (1e-10)
#define MOD 1000000007LL
typedef vector<lint> vec;
typedef vector<vector<lint>> mat;
typedef vector<string> svec;
typedef vector<vector<string>> smat;
template<typename T>inline void numout(T t){bool f=0;for(auto i:t){cout<<(f?" ":"")<<i<INF/2?i:"INF";f=1;}cout<<endl;}
template<typename T>inline void numout2(T t){for(auto i:t)numout(i);}
template<typename T>inline void output(T t){bool f=0;for(auto i:t){cout<<(f?" ":"")<<i;f=1;}cout<<endl;}
template<typename T>inline void output2(T t){for(auto i:t)output(i);}
template<typename T>inline void _output(T t){bool f=0;for(lint i=0;i<t.size();i++){cout<<f?"":" "<<t[i];f=1;}cout<<endl;}
template<typename T>inline void _output2(T t){for(lint i=0;i<t.size();i++)output(t[i]);}
#define rep(i,n) for(lint i=0;i<lint(n);++i)
#define repi(i,a,b) for(lint i=lint(a);i<(lint)(b);++i)
#define rrep(i,n) for(lint i=lint(n)-1;i>=0;--i)
#define rrepi(i,a,b) for(lint i=lint(b)-1;i>=lint(a);--i)
#define irep(i) for(lint i=0;;++i)
#define all(n) begin(n),end(n)
#define dist(a,b,c,d) sqrt(pow(a-c,2)+pow(b-d,2))
inline lint gcd(lint A,lint B){return B?gcd(B,A%B):A;}
inline lint lcm(lint A,lint B){return A/gcd(A,B)*B;}
inline bool chmin(auto& s,const auto& t){bool res=s>t;s=min(s,t);return res;}
inline bool chmax(auto& s,const auto& t){bool res=s<t;s=max(s,t);return res;}
const vector<lint> dx={-1,1,0,0,1,1,-1,-1};
const vector<lint> dy={0,0,-1,1,1,-1,1,-1};
#define SUM(v) accumulate(all(v),0LL)
auto call=[](auto f,auto... args){return f(f,args...);};

class mint {
  using u64 = std::uint_fast64_t;
    public:
    u64 a;
    constexpr mint(const u64 x = 0)noexcept:a(x % MOD){}
    constexpr u64 &value()noexcept{return a;}
    constexpr const u64 &value() const noexcept {return a;}
    constexpr mint operator+(const mint rhs)const noexcept{return mint(*this) += rhs;}
    constexpr mint operator-(const mint rhs)const noexcept{return mint(*this)-=rhs;}
    constexpr mint operator*(const mint rhs) const noexcept {return mint(*this) *= rhs;}
    constexpr mint operator/(const mint rhs) const noexcept {return mint(*this) /= rhs;}
    constexpr mint &operator+=(const mint rhs) noexcept {
        a += rhs.a;
        if (a >= MOD)a -= MOD;
        return *this;
    }
    constexpr mint &operator-=(const mint rhs) noexcept {
        if (a<rhs.a)a += MOD;
        a -= rhs.a;
        return *this;
    }
    constexpr mint &operator*=(const mint rhs) noexcept {
        a = a * rhs.a % MOD;
        return *this;
    }
    constexpr mint operator++(int n) const noexcept {
        return mint(*this) +=1;
    }
    constexpr mint operator--(int n) const noexcept {
        return mint(*this) -=1;
    }
    constexpr mint &operator/=(mint rhs) noexcept {
        u64 exp=MOD-2;
        while (exp) {
            if (exp % 2) {
                *this *= rhs;
            }
            rhs *= rhs;
            exp /= 2;
        }
        return *this;
    }
    mint pow(long long n){
        long long x=a;
        mint ret = 1;
        while(n>0) {
            if(n&1)(ret*=x);
            (x*=x)%=MOD;
            n>>=1;
        }
        return ret;
    }
    mint inv()const{
        int _a=a,b=MOD,u=1,v=0,t;
        while(b>0) {
            t=_a/b;
            swap(_a-=t*b,b);
            swap(u-=t*v,v);
        }
        return mint(u);
    }
    mint comb(lint b){
        using lint=long long;
        static bool init=1;
        static lint fac[1000001],ifac[1000001];
        if(init){
            init=0;
            fac[0]=1;
            ifac[0]=1;
            auto mod_pow=[&](lint x,lint n){
                lint ans = 1;
                while(n != 0){
                    if(n&1)ans=ans*x%MOD;
                    x=x*x%MOD;
                    n=n>>1;
                }
                return ans;
            };
            for(int i=0;i<1000000;i++){
                fac[i+1]=fac[i]*(i+1)%MOD;
                ifac[i+1]=ifac[i]*mod_pow(i+1, MOD-2)%MOD;
            }
        }
        if(a==0&&b==0)return 1;
        if((lint)a<b||a<0)return 0;
        lint tmp=ifac[a-b]*ifac[b]%MOD;
        return tmp*fac[a]%MOD;
    }
    
    friend ostream& operator<<(ostream& lhs, const mint& rhs) noexcept {
        lhs << rhs.a;
        return lhs;
    }
    friend istream& operator>>(istream& lhs,mint& rhs) noexcept {
        lhs >> rhs.a;
        return lhs;
    }
};
inline mint fact(lint a){return a?fact(a-1)*a:1;}

int main(){
    lint n;
    cin>>n;
    n++;
    vec a(n);
    rep(i,n)cin>>a[i];
    set<lint>s;
    lint t=0,u=0;
    rep(i,n){
        if(s.count(a[i])){
            t=i;
        }
        s.insert(a[i]);
    }
    rep(i,n){
        if(s.count(a[t])){
            u=i;
            break;
        }
    }
    repi(i,1,n+1){
        cout<<mint(n).comb(i)-mint(n-(t-u)-1).comb(i-1)<<endl;
    }
}