Problem: You are given a sequence D_1, D_2, ..., D_N of length N. The values of D_i are all distinct. Does a tree with N vertices that satisfies the following conditions exist?

* The vertices are numbered 1,2,..., N.
* The edges are numbered 1,2,..., N-1, and Edge i connects Vertex u_i and v_i.
* For each vertex i, the sum of the distances from i to the other vertices is D_i, assuming that the length of each edge is 1.



If such a tree exists, construct one such tree.

Constraints

* 2 \leq N \leq 100000
* 1 \leq D_i \leq 10^{12}
* D_i are all distinct.

Input

Input is given from Standard Input in the following format:


N
D_1
D_2
:
D_N


Output

If a tree with n vertices that satisfies the conditions does not exist, print `-1`.

If a tree with n vertices that satisfies the conditions exist, print n-1 lines. The i-th line should contain u_i and v_i with a space in between. If there are multiple trees that satisfy the conditions, any such tree will be accepted.

Examples

Input

7
10
15
13
18
11
14
19


Output

1 2
1 3
1 5
3 4
5 6
6 7


Input

2
1
2


Output

-1


Input

15
57
62
47
45
42
74
90
75
54
50
66
63
77
87
51


Output

1 10
1 11
2 8
2 15
3 5
3 9
4 5
4 10
5 15
6 12
6 14
7 13
9 12
11 13

Solution (in CPP):

Solution:
#include<iostream>
#include<algorithm>
#include<map>
using namespace std;

struct node
{
	int f,t;
} t[1000000];

long long a[1000000],size[1000000],id[1000000],hh[1000000],h,tot;
map<long long,int>mp;

void add(long long x,long long y)
{
	t[++h].f=hh[x];
	t[h].t=y;
	hh[x]=h;
}

int cmp(int x,int y)
{
	return a[x]>a[y];
}

void dfs(int x,int y,int z)
{
	tot+=z;
	for(int i=hh[x];i;i=t[i].f)
	{
		int j=t[i].t;
		if(j==y) 
			continue;
		dfs(j,x,z+1);
	}
}

int main()
{
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++) 
	{
		scanf("%lld",a+i);
		mp[a[i]]=i;
		size[i]=1;
		id[i]=i;
	}
	sort(id+1,id+1+n,cmp);
	for(int i=1;i<n;i++)
	{
		long long x=id[i],y=a[x]-(n-size[x])+size[x];
		if(!mp[y]||y>=a[x]) 
		{
			printf("-1\n");
			return 0;
		}
		y=mp[y];
		add(x,y);
		add(y,x);
		size[y]+=size[x];
	}
	dfs(id[n],0,0);
	if(tot!=a[id[n]]) 
	{
		printf("-1\n");
		return 0;
	}
	for(int i=1;i<=n;i++)
		for(int j=hh[i];j;j=t[j].f)
			if(t[j].t>i) 
				printf("%d %d\n",i,t[j].t);
	return 0;
}