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<stdlib.h>
#include<string.h>
#include<math.h>
#define FOR(n) for(int i=0;i<n;i++)
#define FORJ(n) for(int j=0;j<n;j++)
#define PRN(n) printf("%d\n",n)
#define PRF(n) printf("%lf\n",n)
#define PRL(n) printf("%lld\n",n)
#define PRS(s) printf("%s\n",s)
#define PRC(c) printf("%c",c)
#define INF 10000000000
#define mod 1000000007
typedef long long int ll;
int u(const void *a, const void *b){
    return *(ll*)a-*(ll*)b;
}
int d(const void *a, const void *b){
    return *(ll*)b-*(ll*)a;
}
int min(int a,int b){
  if(a>b)return b;
  return a;
}
int max(int a,int b){
  if(a>b)return a;
  return b;
}
int gcd(int a,int b){
  if(!b)return a;
  return gcd(b,a%b);
}
int kt(ll a){
  int sum=0;
  while(a){
    a/=10;
    sum++;
  }
  return sum;
}
int ks(ll a){
  int sum=0;
  while(a){
    sum+=a%10;
    a/=10;
  }
  return sum;
}
int main(void){
  int n;
  scanf("%d",&n);
  int a[n][2];
  FOR(n){
    scanf("%d",&a[i][0]);
    a[i][1]=i+1;
  }
  qsort(a,n,sizeof(int)*2,u);
  FOR(n)PRN(a[i][1]);
  return 0;
}