// 基本テンプレート
 
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <cfloat>
#include <climits>
#include <ctime>
#include <cassert>
#include <numeric>
#include <fstream>
#include <functional>
using namespace std;
 
#define rep(i,a,n) for(int (i)=(a); (i)<(n); (i)++)
#define repq(i,a,n) for(int (i)=(a); (i)<=(n); (i)++)
#define repr(i,a,n) for(int (i)=(a); (i)>=(n); (i)--)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define int long long int
 
template<typename T> void chmax(T &a, T b) {a = max(a, b);}
template<typename T> void chmin(T &a, T b) {a = min(a, b);}
template<typename T> void chadd(T &a, T b) {a = a + b;}
 
typedef pair<int, int> pii;
typedef long long ll;
 
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
const ll INF = 1001001001001001LL;
const ll MOD = 1000000007LL;
 
signed main() {
    int N; cin >> N;
    vector<int> vec(N);
    for(int i=0; i<N; i++) {
        cin >> vec[i];
    }

    vector<int> ans;
    priority_queue<int> pre, suf;
    for(int i=0; i<N; i++) {
        pre.push(vec[i]);
        suf.push(vec[N-i-1]);

        while(pre.size() && suf.size()) {
            if(pre.top() != suf.top()) break;
            pre.pop(); suf.pop();
        }

        if(pre.empty() && suf.empty()) ans.push_back(i+1);
    }

    for(size_t i=0; i<ans.size(); i++) {
        printf("%lld%c", ans[i], (i+1==ans.size() ? '\n' : ' '));
    }
    return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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
// 基本テンプレート
 
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <cfloat>
#include <climits>
#include <ctime>
#include <cassert>
#include <numeric>
#include <fstream>
#include <functional>
using namespace std;
 
#define rep(i,a,n) for(int (i)=(a); (i)<(n); (i)++)
#define repq(i,a,n) for(int (i)=(a); (i)<=(n); (i)++)
#define repr(i,a,n) for(int (i)=(a); (i)>=(n); (i)--)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define int long long int
 
template<typename T> void chmax(T &a, T b) {a = max(a, b);}
template<typename T> void chmin(T &a, T b) {a = min(a, b);}
template<typename T> void chadd(T &a, T b) {a = a + b;}
 
typedef pair<int, int> pii;
typedef long long ll;
 
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
const ll INF = 1001001001001001LL;
const ll MOD = 1000000007LL;
 
signed main() {
    int N; cin >> N;
    vector<int> vec(N);
    for(int i=0; i<N; i++) {
        cin >> vec[i];
    }

    vector<int> ans;
    priority_queue<int> pre, suf;
    for(int i=0; i<N; i++) {
        pre.push(vec[i]);
        suf.push(vec[N-i-1]);

        while(pre.size() && suf.size()) {
            if(pre.top() != suf.top()) break;
            pre.pop(); suf.pop();
        }

        if(pre.empty() && suf.empty()) ans.push_back(i+1);
    }

    for(size_t i=0; i<ans.size(); i++) {
        printf("%lld%c", ans[i], (i+1==ans.size() ? '\n' : ' '));
    }
    return 0;
}

```