#define __USE_MINGW_ANSI_STDIO 0
#include <bits/stdc++.h>

using namespace std;
using ll = long long;
#define int ll
using VI = vector<int>;
using VVI = vector<VI>;
using PII = pair<int, int>;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define PB push_back

const ll LLINF = (1LL<<60);
const int INF = (1LL<<30);
const int MOD = 1000000007;

template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }
template <typename T> bool IN(T a, T b, T x) { return a<=x&&x<b; }
template<typename T> T ceil(T a, T b) { return a/b + !!(a%b); }
template<class S,class T>
ostream &operator <<(ostream& out,const pair<S,T>& a){
  out<<'('<<a.first<<','<<a.second<<')';
  return out;
}
template<class T>
ostream &operator <<(ostream& out,const vector<T>& a){
  out<<'[';
  REP(i, a.size()) {out<<a[i];if(i!=a.size()-1)out<<',';}
  out<<']';
  return out;
}

int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};

int a[100010];
signed main(void)
{
  cin.tie(0);
  ios::sync_with_stdio(false);
	
	int n;
	cin >> n;
	REP(i, n) cin >> a[i];
	
	VI ans, yui(100010, 0);
	map<int, int> kyoko;
	int ptr = -1;
	REP(i, n) {
		kyoko[a[i]]++;
		yui[++ptr] = a[n-i-1];
		while(ptr >= 0) {
			if (kyoko[yui[ptr]]) {
				kyoko[yui[ptr]]--;
				ptr--;
			} else {
				break;
			}
		}
		if(ptr == -1) ans.PB(i+1);
	}
	
	REP(i, ans.size()) {
		cout << ans[i];
		if(i != ans.size()-1) cout << " ";
	}
	cout << 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
#define __USE_MINGW_ANSI_STDIO 0
#include <bits/stdc++.h>

using namespace std;
using ll = long long;
#define int ll
using VI = vector<int>;
using VVI = vector<VI>;
using PII = pair<int, int>;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define PB push_back

const ll LLINF = (1LL<<60);
const int INF = (1LL<<30);
const int MOD = 1000000007;

template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }
template <typename T> bool IN(T a, T b, T x) { return a<=x&&x<b; }
template<typename T> T ceil(T a, T b) { return a/b + !!(a%b); }
template<class S,class T>
ostream &operator <<(ostream& out,const pair<S,T>& a){
  out<<'('<<a.first<<','<<a.second<<')';
  return out;
}
template<class T>
ostream &operator <<(ostream& out,const vector<T>& a){
  out<<'[';
  REP(i, a.size()) {out<<a[i];if(i!=a.size()-1)out<<',';}
  out<<']';
  return out;
}

int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};

int a[100010];
signed main(void)
{
  cin.tie(0);
  ios::sync_with_stdio(false);
	
	int n;
	cin >> n;
	REP(i, n) cin >> a[i];
	
	VI ans, yui(100010, 0);
	map<int, int> kyoko;
	int ptr = -1;
	REP(i, n) {
		kyoko[a[i]]++;
		yui[++ptr] = a[n-i-1];
		while(ptr >= 0) {
			if (kyoko[yui[ptr]]) {
				kyoko[yui[ptr]]--;
				ptr--;
			} else {
				break;
			}
		}
		if(ptr == -1) ans.PB(i+1);
	}
	
	REP(i, ans.size()) {
		cout << ans[i];
		if(i != ans.size()-1) cout << " ";
	}
	cout << endl;
  
  return 0;
}

```