CodeForces 136A Presents

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 from1 ton. 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 contains n space-separated integers: thei-th number isp_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: thei-th number should equal the number of the friend who gave a gift to friend numberi.

Example

Input

42 3 4 1

Output

4 1 2 3

Input

31 3 2

Output

1 3 2

Input

21 2

Output

1 2

题目不难 看懂就可以，将给定的数排序，然后按顺序输出每个数在输入时的序号。

结构体排序就可以。

AC代码：

#include <iostream>#include <stdio.h>#include <math.h>#include <string.h>#include <algorithm>using namespace std;struct shu{    int num,order;}s[105];int cmp(shu a,shu b){    return a.num<b.num;}int main(){    int i,n,num[105],order[105];    while(~scanf("%d",&n))    {        for(i=0;i<n;i++)        {            scanf("%d",&s[i].num);            s[i].order=i;        }        sort(s,s+n,cmp);        for(i=0;i<n-1;i++)        {            printf("%d ",s[i].order+1);        }        printf("%d\n",s[n-1].order+1);    }    return 0;}