Presents（CodeForces 136A）

Presents

Description

Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited n his friends there.

If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to n. Petya remembered that a friend numberi gave a gift to a friend numberp_i. He also remembered that each of his friends received exactly one gift.

Now Petya wants to know for each friend i the number of a friend who has given him a gift.

Input

The first line contains one integer n (1 ≤ n ≤ 100) — the quantity of friends Petya invited to the party. The second line containsn space-separated integers: the i-th number is p_i — the number of a friend who gave a gift to friend numberi. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves.

Output

Print n space-separated integers: the i-th number should equal the number of the friend who gave a gift to friend numberi.

Sample Input

Input

42 3 4 1

Output

4 1 2 3

Input

31 3 2

Output

1 3 2

Input

21 2

Output

1 2

题意：输入一排数，然后将这些数由小到大排序，输出排好序之后数原来的位置
思路：典型的数组利用·，但得注意判断空格。。。。。。
代码：
#include<stdio.h>
#include<string.h>
int main()
{
    int a[200],b[200],n;
    while(~scanf("%d",&n))
    {
        memset(b,0,sizeof(b));  //注意将数组清零，0代表没有，要不然不知道吗个数有没有。。。。。
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            b[a[i]]=i;      
        }
        int h=0;
        for(int i=0;i<105;i++)
        {
            if(b[i]!=0)       //b数组不为零，输出第一个数
            {
                if(h==0)
                {
                    printf("%d",b[i]);
                    h++;   //第一个数之后h就变为了一，就执行else的语句
                }
                else printf(" %d",b[i]);   //输出空格加数
            }
        }
        printf("\n");  //之后换行。。。。。。。
    }
    return 0;
}