Input

The first line of the input contains integer n (2 ≤ n ≤ 100) — the number of cards in the deck. It is guaranteed that n is even.

The second line contains the sequence of n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is equal to the number written on the i-th card.

Output

Print n / 2 pairs of integers, the i-th pair denote the cards that should be given to the i-th player. Each card should be given to exactly one player. Cards are numbered in the order they appear in the input.

It is guaranteed that solution exists. If there are several correct answers, you are allowed to print any of them.

Examples

Input
6
1 5 7 4 4 3
Output
1 3
6 2
4 5
Input
4
10 10 10 10
Output
1 2
3 4

Note

In the first sample, cards are distributed in such a way that each player has the sum of numbers written on his cards equal to 8.

In the second sample, all values ai are equal. Thus, any distribution is acceptable.

題目大意：
給出n個人，每個人的卡片數目，兩個人分成一組，使得每一組的卡片數目和都相等。

思路：

1、貪心的將卡片數排序，從小到大，因為保證有解，那麼最少卡片的將和最多卡片的組隊，第二少卡片的將和第二多卡片的組隊，依次類推。

Ac程式碼：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
    int val,pos;
}a[10000];
int cmp(node a,node b)
{
    return a.val<b.val;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i].val);
            a[i].pos=i+1;
        }
        sort(a,a+n,cmp);
        for(int i=0;i<n/2;i++)
        {
            printf("%d %d\n",a[i].pos,a[n-i-1].pos);
        }
    }
}