Problem Description

You are given an integer sequence of length n+1, a1,a2,…,an+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 109+7.

Note

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≤n≤105
• 1≤ai≤n
• Each of the integers?1,…,n?appears in the sequence.
• n?and?ai?are integers.

Input

Input is given from Standard Input in the following format:

n
a1 a2 ... an+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 109+7.

Example

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 109+7.

題意： 給出 n+1 個(gè)數(shù)，對(duì)于這 n+1?個(gè)整數(shù)，找到長(zhǎng)度從 1~n+1 的不同子序列的個(gè)數(shù)，要求子序列中沒(méi)有相同的數(shù)

思路：容斥原理，首先用求出所有情況，再減去重復(fù)情況即可

Source Program

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #include<bitset> #define EPS 1e-9 #define PI acos(-1.0) #define INF 0x3f3f3f3f #define LL long long const int MOD = 1E9+7; const int N = 100000+5; const int dx[] = {-1,1,0,0,-1,-1,1,1}; const int dy[] = {0,0,-1,1,-1,1,-1,1}; using namespace std;LL a[N]; LL inv[N]; LL fac[N]; map<LL,int> bucket; void init() {fac[0]=1;fac[1]=1;inv[0]=1;inv[1]=1;for(LL i=2; i<N; i++) {fac[i]=fac[i-1]*i%MOD;inv[i]=(MOD-MOD/i)*inv[MOD%i]%MOD;}for(int i=1; i<N; i++)inv[i]=inv[i]*inv[i-1]%MOD; } LL C(LL n, LL m) {if(m>n)return 0;return ((fac[n]*inv[m]%MOD)*inv[n-m])%MOD; } int main() {init();LL n;scanf("%lld",&n);LL pos=0;for(int i=1;i<=n+1;i++){scanf("%lld",&a[i]);if(!bucket[a[i]])bucket[a[i]]=i;elsepos=i;}for(int i=0;i<=n;i++){LL res=0;res=(res+C(n+1,i+1))%MOD;res=(res-C(n-pos+bucket[a[pos]],i))%MOD;while(res<0)res+=MOD;printf("%lld\n",res);}return 0; }

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