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: Accepted)

#include <bits/stdc++.h> // ver2.3.6
#include <boost/math/tools/minima.hpp>
#include <boost/multiprecision/cpp_int.hpp>
#define fmin brent_find_minima
#define int long long
#define endl "\n"
#define ALL(v) (v).begin(),(v).end()
#define COUNT(a,k) upper_bound(ALL(a),k)-lower_bound(ALL(a),k)
#define BIGGER(a,k) a.end()-upper_bound(ALL(a),k)
#define SMALLER(a,k) lower_bound(ALL(a),k)-a.begin()
#define Vi vector<int>
#define VVi vector<Vi>
#define Vs vector<string>
#define Pii pair<int,int>
#define VPii vector<Pii>
#define Tiii tuple<int,int,int>
#define VTiii vector<Tiii>
#define PQi priority_queue<int>
#define PQir priority_queue<int,vector<int>,greater<int>>
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define itos to_string
#define stoi stoll
#define FI first
#define SE second
#define cYES cout<<"YES"<<endl
#define cNO cout<<"NO"<<endl
#define cYes cout<<"Yes"<<endl
#define cNo cout<<"No"<<endl
#define cyes cout<<"yes"<<endl
#define cno cout<<"no"<<endl
#define sortr(v) sort(v,greater<>())
#define rep(i,a,b) for(int i=a;i<b;i++)
#define repeq(i,a,b) for(int i=a;i<=b;i++)
#define repreq(i,a,b) for(int i=a;i>=b;i--)
#define leng(n) (int)(log10(n)+1)
#define dem(a,b) ((a+b-1)/(b))
#define Vin(a) rep(iI,0,a.size())cin>>a[iI]
#define INF 3000000000000000000  //  3.0*10^18(MAXの1/3くらい)
#define MAX LLONG_MAX
#define PI 3.141592653589793238462
#define MOD 1000000007  //  10^9 + 7

using namespace std;
/* fmin(f,L,R,100) で関数の最小値を(x,min)で返す　*/
using boost::math::tools::brent_find_minima;
/* cpp_int (任意の長さのint) */
using boost::multiprecision::cpp_int;

/* デバッグ用 */
void Vout(auto a){if(a.size()==0) cout<<"."<<endl;
else{rep(i,0,a.size()-1)cout<<a[i]<<' ';if(a.size()>0)cout<<a.back()<<endl;}return;}
void Verr(auto a){if(a.size()==0) cerr<<"."<<endl;
else{rep(i,0,a.size()-1)cerr<<a[i]<<' ';if(a.size()>0)cerr<<a.back()<<endl;}return;}
void VVout(auto a){if(a.size()==0)cout<<"."<<endl;
else{rep(i,0,a.size())Vout(a[i]);}return;}
void VVerr(auto a){if(a.size()==0)cerr<<"."<<endl;
else{rep(i,0,a.size())Verr(a[i]);}return;}
void VPout(auto a){if(a.size()==0)cout<<"."<<endl;
else{rep(i,0,a.size())cout<<a[i].FI<<' '<<a[i].SE<<endl;}return;}
void VPerr(auto a){if(a.size()==0)cerr<<"."<<endl;
else{rep(i,0,a.size())cerr<<a[i].FI<<' '<<a[i].SE<<endl;}return;}

/* 便利関数 */
int gcd(int a,int b){return b?gcd(b,a%b):a;} // 最大公約数gcd
int lcm(int a,int b){return a/gcd(a,b)*b;} // 最小公倍数lcm
int mypow(int x, int n, int m){ // 累乗x^n(mod m) O(log(n))
if(n==0)return 1;if(n%2==0)return mypow(x*x%m,n/2,m);else return x*mypow(x,n-1,m)%m;}
int scomb(int n, int r){if((n-r)<r)r=n-r; // nCr(小さい場合)
int a=1;for(int i=n;i>n-r;--i){a=a*i;}for(int i=1;i<r+1;++i){a=a/i;}return a;}
int comb(int n, int r){if((n-r)<r)r=n-r; // nCr(%MOD)
int a=1;for(int i=n;i>n-r;--i){a=a*i%MOD;}for(int i=1;
i<r+1;++i){a=a*mypow(i,MOD-2,MOD)%MOD;}return a%MOD;}
Vi stpowV(){Vi a(100001);
a[0]=1;repeq(i,1,100000)a[i]=a[i-1]*i%MOD;return a;}
//Vi stpow = stpowV();  // 階乗配列(%MOD)
void press(auto &v){v.erase(unique(ALL(v)),v.end());} // 圧縮
Vi Vsum(Vi &v){Vi S(v.size()+1);rep(i,1,S.size())S[i]+=v[i-1]+S[i-1];return S;}// 累積和
int keta(int k,int i){string s = itos(k);return s[s.size()-i]-'0';} // i桁目の数字

struct mint { // 出力するときだけ.xをつける
  int x;
  mint(int x=0):x((x%MOD+MOD)%MOD){}
  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(int 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;
  }
};


int n;
int x,y,z;
vector<mint> xyzc,xyc;


int xyzCi(int i){
  if(i < 0 || i > x+y+z) return 0;
  return xyzc[i].x;
}
int xyCi(int i){
  if(i < 0 || i > x+z) return 0;
  return xyc[i].x;
}



signed main() {
  cin.tie(0);cout.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(12);
  
  cin >> n;
  n++;
  Vi a(n);
  Vin(a);
  
  map<int,int> kind;
  
  rep(i,0,n){
    if(kind[a[i]] == 0) kind[a[i]] = i+1;
    else {
      x = kind[a[i]]-1;
      y = i;
    }
  }
  
  z = n-y-1;
  y = y-(x+1);
  
  xyzc = vector<mint>(x+y+z+1);
  xyc = vector<mint>(x+z+1);
  
  xyzc[0] = 1;
  
  repeq(i,1,x+y+z){
    xyzc[i] = xyzc[i-1]*(x+y+z+1-i);
    xyzc[i] /= i;
  }
  
  xyc[0] = 1;
  
  repeq(i,1,x+z){
    xyc[i] = xyc[i-1]*(x+z+1-i);
    xyc[i] /= i;
  }
  
  mint ans;
  
  repeq(i,1,n){
    ans = 0;
    ans += xyzCi(i);
    ans += xyzCi(i-2);
    ans += xyzCi(i-1)*2;
    ans -= xyCi(i-1);
    
    cout << ans.x << endl;
  }
  
  return 0;
}