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 from 1 to n. Petya remembered that a friend number i gave a gift to a friend number pi. 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: the i-th number is pi — the number of a friend who gave a gift to friend number i. 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 number i.

    Examples

Input

4
2 3 4 1

Output

4 1 2 3

Input

3
1 3 2

Output

1 3 2

Input

2
1 2

Output

1 2

题目大意
一开始我很懵逼啊啊啊，然后再纸上划拉了一下，特别绕的解释是
输入部分：第i个数表示第i个人的礼物给了谁
输出部分：第i个数表示谁的礼物给了第i 个人
是不是特别绕啊
划拉划拉就发现用两个数组就可以啊
比如
2 3 4 1
利用另一个数组
b[2]=1
b[3]=2
b[4]=3
b[1]=4然后用顺序输出就好啦

代码如下

#include<stdio.h>
#include<string.h>
int main()
{
 int n,i,j;
 int a[110],b[110];
 memset(b,0,sizeof(b));
 scanf("%d",&n);
 for(i=1;i<=n;i++)
 {
     scanf("%d",&a[i]);
     b[a[i]]=i;
    }
    for(i=1;i<=n;i++)
    {
     if(i==1)
        printf("%d",b[i]);
     else
        printf(" %d",b[i]);
    }
    printf("\n");
    return 0;  
}