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 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
 
int main(void){
    long n,i,j;
    long e1,e2,e3,e4,e5,e6,e7;
    long a[100000];
    long b[100000];
    long center;
    scanf("%ld",&n);
  
    e1=n/8;
    e2=n*2/8;
    e3=n*3/8;
    e4=n*4/8;
    e5=n*5/8;
    e6=n*6/8;
    e7=n*7/8;
  
    for(i=0; i<=n-1; i++){
        scanf("%ld ",&a[i]);
    }
    for(i=0; i<=n-1; i++){
        for(j=0; j<=n-1; j++){   
            if(a[j]==i+1){
               b[i] = j+1;
               break;
            }
            if(a[e1+j]==i+1){
               b[i] = e1+j+1;
               break;
            }
             if(a[e2+j]==i+1){
               b[i] = e2+j+1;
               break;
            }
            if(a[e3+j]==i+1){
               b[i] = e3+j+1;
               break;
            }
             if(a[e4+j]==i+1){
               b[i] = e4+j+1;
               break;
            }
            if(a[e5+j]==i+1){
               b[i] = e5+j+1;
               break;
            }
            if(a[e6+j]==i+1){
               b[i] = e6+j+1;
               break;
            }
            if(a[e7+j]==i+1){
               b[i] = e7+j+1;
               break;
            }
        }
    }
    for(i=0; i<=n-1; i++) printf("%ld ",b[i]);
  
    return 0;
}