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: Accepted)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#define MAX(X, Y) (((X) > (Y))? (X) : (Y))
#define MIN(X, Y) (((X) < (Y))? (X) : (Y))
int compare_int(const void *a, const void *b){
  return *(long int*)b - *(long int*)a; // 大きい順
}
int compare_int_small(const void *a, const void *b){
  return *(long int*)a - *(long int*)b; // 小さい順
}
// 公約数
long int gcd(long int a, long int b){
  if(b == 0){
    return a;
  }else{
    return gcd(b, a % b);
  }
}
// 最小公倍数 = a * b / 最大公約数
// 公倍数
long int lcm(long int a, long int b){
  return a * b / gcd(a, b);
}



// min
int main(void){

  long int n;
  long int a[100001];
  long int tmp;

  long int ans = 0;;

  scanf("%ld", &n);
  
  for(long int i = 1; i <= n; i++){
    scanf("%ld", &tmp);
    a[tmp] = i;
  }

  //  qsort(a, n, sizeof(long int), compare_int_small);

  //  fprintf(stderr, "n / 2 = %f\n", 3.0/5.0);
  /*
  for(long int i = 1; i <= n; i++){
    for(long int j = 1; j <= n; j++){
      if(i == a[j]){
	printf("%ld\n", j);
	break;
      }
    }
  }
  */
  for(long int i = 1; i <= n; i++){
    printf("%ld\n", a[i]);
  }


  return 0;
}