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++14 (Clang 3.8.0) (judge verdict: Accepted)

#include <algorithm>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <functional>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <vector>
using namespace std;

typedef long long ll;
typedef pair<int, int> P;

const int dx[4] = {-1, 0, 0, 1};
const int dy[4] = {0, -1, 1, 0};

// Self settings
// clang-format off
#define MAX_N 100000
#define REP(i, N) for (int i = 0; i < (int)(N); ++i)
#define SLN(i,N) (i == N-1 ? "\n" : " ")
ll fact(ll n) { ll res = 1; for(ll i=2;i<=n;++i) res = res * i; return res;}
ll nCr(ll n, ll r) {return (fact(n)/fact(n-r)*fact(r)) ;}
ll gcd(ll a,ll b){if(b==0)return a;return gcd(b,a%b);}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
const ll MOD = 1e9+7;
const ll INF = 1LL << 60;
const int inf = 1 << 31;
// clang-format on

int N;
int A[MAX_N];

int main(void)
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> N;
  priority_queue<P, vector<P>, greater<P>> q;
  REP(i, N)
  {
    int x;
    cin >> x;
    q.push(P(x, i + 1));
  }

  int i = 0;
  while (!q.empty()) {
    P p = q.top();
    q.pop();
    cout << p.second << SLN(i, N);
    i++;
  }
}