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>
using namespace std;
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
typedef long long LL;
typedef unsigned long long uLL;
typedef pair<int,int> pii;

const int inf=0x3f3f3f3f;
const LL Linf=1ll<<61;

const int maxn=100111;
int n,pa[maxn],sz[maxn];
LL d[maxn];
pair<LL,int> vs[maxn];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%lld",d+i);
		vs[i]=mp(d[i],i);
		sz[i]=1;
	}
	sort(vs+1,vs+n+1);for(int i=1;i<=n;i++)d[i]=vs[i].ff;
	for(int i=n;i>=2;i--)
	{
		int dlt=(n-sz[i])-sz[i],id=lower_bound(d+1,d+n+1,d[i]-dlt)-d;
		if(id>=i||d[id]!=d[i]-dlt)
		{
			puts("-1");
			return 0;
		}
		pa[i]=id;
		sz[id]+=sz[i];
	}
	LL sum=0;
	for(int i=1;i<=n;i++)sum+=sz[i]-1;
	if(sum!=d[1])
	{
		puts("-1");
		return 0;
	}
	for(int i=2;i<=n;i++)printf("%d %d\n",vs[pa[i]].ss,vs[i].ss);
	return 0;
}