cf701A Cards

There are n cards (n is even) in the deck. Each card has a positive integer written on it. n / 2 people will play new card game. At the beginning of the game each player gets two cards, each card is given to exactly one player.

Find the way to distribute cards such that the sum of values written of the cards will be equal for each player. It is guaranteed that it is always possible.

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.

水题简直了

排个序然后从头尾各取一个

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #include<deque>
 #include<set>
 #include<map>
 #include<ctime>
 #define LL long long
 #define inf 0x7ffffff
 #define pa pair<int,int>
 #define pi 3.1415926535897932384626433832795028841971
 using namespace std;
 inline LL read()
 {
     LL x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 inline void write(LL a)
 {
     if (a<){printf("-");a=-a;}
     if (a>=)write(a/);
     putchar(a%+'');
 }
 inline void writeln(LL a){write(a);printf("\n");}
 struct e{int x,y;}a[];
 bool operator <(e a,e b)
 {return a.x<b.x;}
 int n,tot;
 int main()
 {
     n=read();
     for (int i=;i<=n;i++)a[i].x=read(),tot+=a[i].x,a[i].y=i;
     tot/=n;
     sort(a+,a+n+);
     for (int i=;i<=n/;i++)
     {
         printf("%d %d\n",a[i].y,a[n-i+].y);
     }
 }

cf701A