Problem:
Score : 300 points

Problem Statement

Takahashi is a teacher responsible for a class of N students.

The students are given distinct student numbers from 1 to N.

Today, all the students entered the classroom at different times.

According to Takahashi's record, there were A_i students in the classroom when student number i entered the classroom (including student number i).

From these records, reconstruct the order in which the students entered the classroom.

Constraints

1 \le N \le 10^5

1 \le A_i \le N

A_i \neq A_j (i \neq j)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \ldots A_N

Output

Print the student numbers of the students in the order the students entered the classroom.

Sample Input 1

3
2 3 1

Sample Output 1

3 1 2

First, student number 3 entered the classroom.

Then, student number 1 entered the classroom.

Finally, student number 2 entered the classroom.

Sample Input 2

5
1 2 3 4 5

Sample Output 2

1 2 3 4 5

Sample Input 3

8
8 2 7 3 4 5 6 1

Sample Output 3

8 2 4 5 6 7 3 1

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
typedef long long int ll;
typedef long double ld;
using namespace std;
const long long int INF = 1e18;
const long long int MOD = 1e9+7;

typedef pair<ll, ll> pairs;
typedef vector<pairs> p;

struct edge{
    ll to, cost;
};


ll gcd(ll a,ll b){
    if(b == 0){
        return a;
    }else{
        return gcd(b, a%b);
    }
}

ll lcm(ll a, ll b){
    return a / gcd(a, b) * b;
}

ll keta(ll N){
    int tmp{};
    while( N > 0 ){
        tmp += ( N % 10 );
        N /= 10;
    }
    N = tmp;
    return N;
}

// 素数
vector<bool> IsPrime;
void sieve(size_t max){
    if(max+1 > IsPrime.size()){
        IsPrime.resize(max+1,true);
    } 
    IsPrime[0] = false;
    IsPrime[1] = false;

    for(size_t i=2; i*i<=max; ++i){
        if(IsPrime[i]){
            for(size_t j=2; i*j<=max; ++j){
                IsPrime[i*j] = false;
            }
        }
    }
}

// べき乗の余り
ll modpow(ll a, ll n){
    ll res = 1;
    while(n){
        if(n & 1){
            res = res * a % MOD;
        }
        a = a * a % MOD;
        n = n >> 1;
    }
    return res;
}

// コンビネーション
ll MAXN = 2e5+10;
vector<ll> inv(MAXN);
ll nCk(ll n, ll k) {
	ll res = 1;
	for(ll i = 1; i <= k; i++) {
		res = res*(n-i+1) % MOD;
	}
 
	for(ll i = 1; i <= k; i++) {
		res = res*inv[i] % MOD;
	}
	return res;
}

// 回文
bool kai(string S){
    bool flag = true;
    for(ll i=0;i<S.size()/2;++i){
        if(S[i] != S[S.size()-i-1]){
            flag = false;
            break;
        }
    }
    return flag;
}
// ---------------------------------------------

int main(){
    ll n;
    cin>>n;
    vector<pair<ll, ll>> a(n);
    for(ll i=0;i<n;++i){
        ll b;
        cin>>b;
        a[i] = make_pair(b, i+1);
    }
    sort(a.begin(), a.end());

    for(ll i=0;i<n;++i){
        if(i != 0){
            cout<<' ';
        }
        cout<<a[i].second;
    }
    cout<<endl;
    return 0;
}