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
#include<bits/stdc++.h>
#define LL long long
using namespace std;
struct var{
	int fr,to,nxt;
}edge[101001];
int n,size[101001],cnt,head[101001];
LL d[101001],dis[101001];
map<LL,int>num;
void add(int u,int v)
{
	cnt++;
	edge[cnt].to=v;
	edge[cnt].fr=u;
	edge[cnt].nxt=head[u];
	head[u]=cnt;
}
void dfs(int o,int f)
{
	for(int i=head[o];i;i=edge[i].nxt)
	{
		int v=edge[i].to;
		if(v==f)continue;
		dis[v]=dis[o]+1;
		dfs(v,o);
	}
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%lld",&d[i]);
		num[d[i]]=i;
		size[i]=1;
	}
	sort(d+1,d+n+1);
	for(int i=n;i>1;i--)
	{
		int u=num[d[i]];
		LL tmp=n-2*size[u],f=d[i]-tmp;
		if(tmp<0)
		{
			printf("-1");
			return 0;
		}
		if(!num.count(f))
		{
			printf("-1");
			return 0;
		}
		int x=num[f];
		add(x,u);
		size[x]+=size[u];
	}
	dfs(num[d[1]],0);
	LL sum=0;
	for(int i=1;i<=n;i++)sum+=dis[i];
	if(sum!=d[1])
	{
		printf("-1");
		return 0;
	}
	for(int i=1;i<=cnt;i++)
	{
		printf("%d %d\n",edge[i].fr,edge[i].to);
	}
}