#include <bits/stdc++.h>

#define _overload(_1,_2,_3,name,...) name
#define _rep(i,n) _range(i,0,n)
#define _range(i,a,b) for(int i=int(a);i<int(b);++i)
#define rep(...) _overload(__VA_ARGS__,_range,_rep,)(__VA_ARGS__)

#define _rrep(i,n) _rrange(i,n,0)
#define _rrange(i,a,b) for(int i=int(a)-1;i>=int(b);--i)
#define rrep(...) _overload(__VA_ARGS__,_rrange,_rrep,)(__VA_ARGS__)

#define _all(arg) begin(arg),end(arg)
#define uniq(arg) sort(_all(arg)),(arg).erase(unique(_all(arg)),end(arg))
#define getidx(ary,key) lower_bound(_all(ary),key)-begin(ary)
#define clr(a,b) memset((a),(b),sizeof(a))
#define bit(n) (1LL<<(n))
#define popcount(n) (__builtin_popcountll(n))

using namespace std;

template<class T>bool chmax(T &a, const T &b) { return (a < b) ? (a = b, 1) : 0;}
template<class T>bool chmin(T &a, const T &b) { return (b < a) ? (a = b, 1) : 0;}

using ll = long long;
using R = long double;
const R EPS = 1e-9L; // [-1000,1000]->EPS=1e-8 [-10000,10000]->EPS=1e-7
inline int sgn(const R& r) {return (r > EPS) - (r < -EPS);}
inline R sq(R x) {return sqrt(max(x, 0.0L));}

const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};

// Problem Specific Parameter:

const int limit = 100010;
int a[limit];
multiset<int> ord, rev;


int main(void) {
	int n;
	cin >> n;
	rep(i, n) cin >> a[i];


	vector<int> ans;

	rep(i, n) {
		const int ridx = n - 1 - i;
		
		{
			auto it = rev.find(a[i]);
			if(it != end(rev)){
				rev.erase(it);
			}else{
				ord.insert(a[i]);
			}
		}

		{
			auto it = ord.find(a[ridx]);
			if(it != end(ord)){
				ord.erase(it);
			}else{
				rev.insert(a[ridx]);
			}
		}

		if(ord.empty() and rev.empty()) ans.push_back(i + 1);
	}

	const int m = ans.size();
	rep(i,m) cout << (i?" ":"") << ans[i];
	cout << endl;


	return 0;
}

### Prompt

Create a solution in CPP for 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>

#define _overload(_1,_2,_3,name,...) name
#define _rep(i,n) _range(i,0,n)
#define _range(i,a,b) for(int i=int(a);i<int(b);++i)
#define rep(...) _overload(__VA_ARGS__,_range,_rep,)(__VA_ARGS__)

#define _rrep(i,n) _rrange(i,n,0)
#define _rrange(i,a,b) for(int i=int(a)-1;i>=int(b);--i)
#define rrep(...) _overload(__VA_ARGS__,_rrange,_rrep,)(__VA_ARGS__)

#define _all(arg) begin(arg),end(arg)
#define uniq(arg) sort(_all(arg)),(arg).erase(unique(_all(arg)),end(arg))
#define getidx(ary,key) lower_bound(_all(ary),key)-begin(ary)
#define clr(a,b) memset((a),(b),sizeof(a))
#define bit(n) (1LL<<(n))
#define popcount(n) (__builtin_popcountll(n))

using namespace std;

template<class T>bool chmax(T &a, const T &b) { return (a < b) ? (a = b, 1) : 0;}
template<class T>bool chmin(T &a, const T &b) { return (b < a) ? (a = b, 1) : 0;}

using ll = long long;
using R = long double;
const R EPS = 1e-9L; // [-1000,1000]->EPS=1e-8 [-10000,10000]->EPS=1e-7
inline int sgn(const R& r) {return (r > EPS) - (r < -EPS);}
inline R sq(R x) {return sqrt(max(x, 0.0L));}

const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};

// Problem Specific Parameter:

const int limit = 100010;
int a[limit];
multiset<int> ord, rev;


int main(void) {
	int n;
	cin >> n;
	rep(i, n) cin >> a[i];


	vector<int> ans;

	rep(i, n) {
		const int ridx = n - 1 - i;
		
		{
			auto it = rev.find(a[i]);
			if(it != end(rev)){
				rev.erase(it);
			}else{
				ord.insert(a[i]);
			}
		}

		{
			auto it = ord.find(a[ridx]);
			if(it != end(ord)){
				ord.erase(it);
			}else{
				rev.insert(a[ridx]);
			}
		}

		if(ord.empty() and rev.empty()) ans.push_back(i + 1);
	}

	const int m = ans.size();
	rep(i,m) cout << (i?" ":"") << ans[i];
	cout << endl;


	return 0;
}

```