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>
using namespace std;

#define VA_NUM_ARGS(...) VA_NUM_ARGS_IMPL_((0,__VA_ARGS__, 6,5,4,3,2,1))
#define VA_NUM_ARGS_IMPL_(tuple) VA_NUM_ARGS_IMPL tuple
#define VA_NUM_ARGS_IMPL(_0,_1,_2,_3,_4,_5,_6,N,...) N
#define macro_dispatcher(macro, ...) macro_dispatcher_(macro, VA_NUM_ARGS(__VA_ARGS__))
#define macro_dispatcher_(macro, nargs) macro_dispatcher__(macro, nargs)
#define macro_dispatcher__(macro, nargs) macro_dispatcher___(macro, nargs)
#define macro_dispatcher___(macro, nargs) macro ## nargs
#define Debug1(a)           cout<<#a<<"="<<(a)<<"\n"
#define Debug2(a,b)         cout<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<"\n"
#define Debug3(a,b,c)       cout<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<"\n"
#define Debug4(a,b,c,d)     cout<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<"\n"
#define Debug5(a,b,c,d,e)   cout<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<", "<<#e<<"="<<(e)<<"\n"
#define Debug6(a,b,c,d,e,f) cout<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<", "<<#e<<"="<<(e)<<", "<<#f<<"="<<(f)<<"\n"
#define Debug(...) macro_dispatcher(Debug, __VA_ARGS__)(__VA_ARGS__)
#define DA(a,n) cout<<#a<<"=["; printarray(a,n); cout<<"]\n"
#define DAR(a,n,s) cout<<#a<<"["<<s<<"-"<<n-1<<"]=["; printarray(a,n,s); cout<<"]\n"

#define TT1 template<class T>
#define TT1T2 template<class T1, class T2>
#define TT1T2T3 template<class T1, class T2, class T3>
template<class T, size_t N> ostream& operator << (ostream& os, const array<T, N>& v);
TT1T2 ostream& operator << (ostream& os, const pair<T1, T2>& p){ return os <<"("<<p.first<<", "<< p.second<<")"; }
TT1 ostream& operator << (ostream& os, const vector<T>& v){       bool f=1;os<<"[";for(auto& i : v) { if (!f)os << ", ";os<<i;f=0;}return os << "]"; }
template<class T, size_t N> ostream& operator << (ostream& os, const array<T, N>& v) {     bool f=1;os<<"[";for(auto& i : v) { if (!f)os << ", ";os<<i;f=0;}return os << "]"; }
TT1T2 ostream& operator << (ostream& os, const set<T1, T2>&v){    bool f=1;os<<"[";for(auto& i : v) { if (!f)os << ", ";os<<i;f=0;}return os << "]"; }
TT1T2 ostream& operator << (ostream& os, const multiset<T1,T2>&v){bool f=1;os<<"[";for(auto& i : v) { if (!f)os << ", ";os<<i;f=0;}return os << "]"; }
TT1T2T3 ostream& operator << (ostream& os, const map<T1,T2,T3>& v){ bool f = 1; os << "["; for (auto& ii : v) { if (!f)os << ", "; os << "(" << ii.first << " -> " << ii.second << ") "; f = 0; }return os << "]"; }
TT1T2 ostream& operator << (ostream& os, const multimap<T1, T2>& v){ bool f = 1; os << "["; for (auto& ii : v) { if (!f)os << ", "; os << "(" << ii.first << " -> " << ii.second << ") "; f = 0; }return os << "]"; }
TT1T2 ostream& operator << (ostream& os, priority_queue<T1, T2> v) { bool f = 1; os << "["; while (!v.empty()) { auto x = v.top(); v.pop(); if (!f) os << ", "; f = 0; os << x; } return os << "]"; }
TT1T2 void printarray(const T1& a, T2 sz, T2 beg = 0){ for (T2 i = beg; i<sz; i++) cout << a[i] << " "; cout << endl; }

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> Pi;
typedef pair<ll, ll> Pll;

#define Fi first
#define Se second
#define pb(x) push_back(x)
#define sz(x) (int)x.size()
#define rep(i, n) for(int i=0;i<n;i++)
#define repp(i, n) for(int i=1;i<=n;i++)
#define all(x) x.begin(), x.end()
#define geti1(X) cin >> X
#define geti2(X,Y) cin >> X >> Y
#define geti3(X,Y,Z) cin >> X >> Y >> Z
#define geti4(X,Y,Z,W) cin >> X >> Y >> Z >> W
#define GET_MACRO(_1,_2,_3,_4,NAME,...) NAME
#define geti(...) GET_MACRO(__VA_ARGS__, geti4, geti3, geti2, geti1) (__VA_ARGS__)

#define INF 987654321

#define endl '\n'
const ll mod = 1e9+7;
ll fpow(ll a, ll b){
	ll res = 1, t = a;
	while( b ){
		if( b&1 ) res = (res*t)%mod;
		t = (t*t)%mod;
		b >>= 1;
	}
	return res;
}
ll inv(ll x){ return fpow(x,mod-2); }
int p[100500];
ll fact[100500];
ll comb(ll n, ll m){
	if( m > n ) return 0LL;
	if( m == n || m == 0 ) return 1LL;
	ll res = fact[n];
	res = (res * inv(fact[m]))%mod;
	res = (res * inv(fact[n-m]))%mod;
	//if( n == 0 ) Debug(m,res);
	return res;
}
ll ans[100500];
int N;
int main() {
	fact[0] = 1;
	for(ll i=1;i<=100200;i++) fact[i] = (fact[i-1]*i)%mod;

	scanf("%d",&N);
	repp(i,N+1) scanf("%d",&p[i]);
	map<int,vector<int>> mp;
	repp(i,N+1) mp[p[i]].pb(i);
	int i=0,j=0;
	for(auto e : mp){
		if( sz(e.Se) >= 2 ){
			i = min(e.Se[0],e.Se[1]);
			j = max(e.Se[0],e.Se[1]);
		}
	}

	for(ll i=1;i<=N;i++){
		ans[i] = (ans[i] + comb(N,i))%mod;
	}
	ll l = j-i;
	for(ll m=1;m<=N;m++){
		ll tmp = comb(N,m-1);
		tmp = (tmp - comb(N-l,m-1) + mod)%mod;
		ans[m] = (ans[m] + tmp)%mod;
	}


	for(ll i=1;i<=N;i++) printf("%lld\n",ans[i]);
	printf("1\n");
}