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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <limits.h>
#include <stdlib.h>
#include <math.h>


long int merge(long int a1[], long int a2[], long int b1[], long int b2[], long int a1_size, long int a2_size, long int b1_size, long int b2_size, long int a[], long int b[]);
long int mergeSort(long int a[], long int a_size, long int b[], long int b_size);

int main(void)
{
	long int n;
	long int i, j;
	long int a[200000];
	long int b[200000];
	
	scanf("%ld", &n);
	
	for(i = 0; i < n; i++) {
		scanf("%ld", &a[i]);
		b[i] = i + 1;
	}
	
	mergeSort(a, n, b, n);
	
	for(i = 0; i < n - 1; i++) {
		printf("%ld ", b[i]);
	}
	printf("%ld", b[n - 1]);
	
	return 0;
}

long int merge(long int a1[], long int a2[], long int b1[], long int b2[], long int a1_size, long int a2_size, long int b1_size, long int b2_size, long int a[], long int b[])
{
	long int i = 0, j = 0;
	
	while( (i < a1_size ) || (j < a2_size) ) {
		if( (j >= a2_size ) || ( (i < a1_size) && (a1[i] < a2[j]) ) ) {
			a[i + j] = a1[i];
			b[i + j] = b1[i];
			i++;
		} else {
			if( (a1[i] == a2[j]) && (b1[i] > b2[j]) ) {
				a[i + j]  = a1[i];
				b[i + j]  = b1[i];
				i++;
			} else {
				a[i + j] = a2[j];
				b[i + j] = b2[j];
				j++;
			}
		}
	}
	
	return 0;
}

long int mergeSort(long int a[], long int a_size, long int b[], long int b_size)
{
	long int m = a_size / 2, n = a_size - m;
	long int a1[m];
	long int a2[n];
	long int b1[m];
	long int b2[n];
	long int i, j;
	
	if(a_size > 1) {
		for(i = 0; i < m; i++) {
			a1[i] = a[i];
			b1[i] = b[i];
		}
		for(i = 0; i < n; i++) {
			a2[i] = a[m + i];
			b2[i] = b[m + i];
		}
		mergeSort(a1, m, b1, m);
		mergeSort(a2, n, b2, n);
		merge(a1, a2, b1, b2, m, n, m, n, a, b);
	}
	
	return 0;
}