#include <bits/stdc++.h>
using namespace std;
#define int long long
#define stoi stoll
using ll=long long;
using vi=vector<int>;
using pii=pair<int,int>;
#define ALL(c) begin(c),end(c)
#define RALL(c) rbegin(c),rend(c)
#define ITR(i,b,e) for(auto i=(b);i!=(e);++i)
#define FORE(x,c) for(auto &x:c)
#define REPF(i,a,n) for(int i=a,i##len=(int)(n);i<i##len;++i)
#define REP(i,n) REPF(i,0,n)
#define REPR(i,n) for(int i=(int)(n);i>=0;--i)
#define SZ(c) ((int)c.size())
#define CONTAIN(c,x) (c.find(x)!=end(c))
#define OUTOFRANGE(y,x,h,w) ((y)<0||(x)<0||(y)>=(h)||(x)>=(w))
#define dump(...)
const signed INF_=1001001001; const long long INF=1001001001001001001LL;
const int DX[9]={0,1,0,-1,1,1,-1,-1,0},DY[9]={-1,0,1,0,-1,1,1,-1,0};
template<class T> ostream& operator<<(ostream &os,const vector<T> &v) {
    ITR(i,begin(v),end(v))os<<*i<<(i==end(v)-1?"":" ");return os;}
template<class T> istream& operator>>(istream &is,vector<T> &v) {
    ITR(i,begin(v),end(v)) is>>*i;return is;}
template<class T,class U> istream& operator>>(istream &is, pair<T,U> &p) {
    is>>p.first>>p.second;return is;}
template<class T> bool chmax(T &a,const T &b){if(a<b){a=b;return 1;}return 0;}
template<class T> bool chmin(T &a,const T &b){if(a>b){a=b;return 1;}return 0;}
template<class T> using heap=priority_queue<T,vector<T>,greater<T>>;
struct{template<class T> operator T(){T x;cin>>x;return x;}} IN;
struct before_main_function {
    before_main_function() {
        cin.tie(0);ios::sync_with_stdio(false);
        cout<<setprecision(15)<<fixed;
        #define endl "\n"
    }
} before_main_function;
//------------------8<------------------------------------8<--------------------

signed main() {
    int n;
    cin>>n;
    vector<int> a(n);
    cin>>a;
    map<int,int> f,b;
    vector<int> ans;
    REP(i,n) {
        f[a[i]]++;
        b[a[n-1-i]]++;
        if(f==b) {
            ans.push_back(i+1);
        }
    }
    cout<<ans<<endl;
    return 0;
}



### Prompt

In CPP, your task is to solve the following problem:

problem

Given a sequence $ a_i $ of length $ N $. Output all integers $ K (1 \ le K \ le N) $ that satisfy the following conditions.

Condition: Well sorted $ a_1, \ cdots, a_K $ matches $ a_ {N-K + 1}, \ cdots, a_N $.





Example

Input

8
5 2 4 9 4 9 2 5


Output

1 2 4 6 7 8

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define stoi stoll
using ll=long long;
using vi=vector<int>;
using pii=pair<int,int>;
#define ALL(c) begin(c),end(c)
#define RALL(c) rbegin(c),rend(c)
#define ITR(i,b,e) for(auto i=(b);i!=(e);++i)
#define FORE(x,c) for(auto &x:c)
#define REPF(i,a,n) for(int i=a,i##len=(int)(n);i<i##len;++i)
#define REP(i,n) REPF(i,0,n)
#define REPR(i,n) for(int i=(int)(n);i>=0;--i)
#define SZ(c) ((int)c.size())
#define CONTAIN(c,x) (c.find(x)!=end(c))
#define OUTOFRANGE(y,x,h,w) ((y)<0||(x)<0||(y)>=(h)||(x)>=(w))
#define dump(...)
const signed INF_=1001001001; const long long INF=1001001001001001001LL;
const int DX[9]={0,1,0,-1,1,1,-1,-1,0},DY[9]={-1,0,1,0,-1,1,1,-1,0};
template<class T> ostream& operator<<(ostream &os,const vector<T> &v) {
    ITR(i,begin(v),end(v))os<<*i<<(i==end(v)-1?"":" ");return os;}
template<class T> istream& operator>>(istream &is,vector<T> &v) {
    ITR(i,begin(v),end(v)) is>>*i;return is;}
template<class T,class U> istream& operator>>(istream &is, pair<T,U> &p) {
    is>>p.first>>p.second;return is;}
template<class T> bool chmax(T &a,const T &b){if(a<b){a=b;return 1;}return 0;}
template<class T> bool chmin(T &a,const T &b){if(a>b){a=b;return 1;}return 0;}
template<class T> using heap=priority_queue<T,vector<T>,greater<T>>;
struct{template<class T> operator T(){T x;cin>>x;return x;}} IN;
struct before_main_function {
    before_main_function() {
        cin.tie(0);ios::sync_with_stdio(false);
        cout<<setprecision(15)<<fixed;
        #define endl "\n"
    }
} before_main_function;
//------------------8<------------------------------------8<--------------------

signed main() {
    int n;
    cin>>n;
    vector<int> a(n);
    cin>>a;
    map<int,int> f,b;
    vector<int> ans;
    REP(i,n) {
        f[a[i]]++;
        b[a[n-1-i]]++;
        if(f==b) {
            ans.push_back(i+1);
        }
    }
    cout<<ans<<endl;
    return 0;
}



```