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;
const int maxn=1E5+5;
typedef long long int ll;
int n,size,head[maxn],ans[maxn];
ll a[maxn],sum[maxn];
map<ll,ll>what;
struct edge
{
	int to,next;
}E[maxn*2];
inline void add(int u,int v)
{
	E[++size].to=v;
	E[size].next=head[u];
	head[u]=size;
}
ll dfs(int u,int F,ll d)
{
	ll sum=0;
	for(int i=head[u];i;i=E[i].next)
	{
		int v=E[i].to;
		if(v==F)
			continue;
		ans[v]=u;
		sum+=dfs(v,u,d+1);
	}
	return sum+d;
}
inline void solve()
{
	for(int i=1;i<=n;++i)
		sum[i]=1;
	for(int i=1;i<n;++i)
	{
		ll x=a[i]-n+sum[what[a[i]]]*2;
		if(!what[x])
		{
			cout<<-1<<endl;
			exit(0);
		}
		add(what[x],what[a[i]]);
		sum[what[x]]+=sum[what[a[i]]];
	}
}
int main()
{
	ios::sync_with_stdio(false);
	cin>>n;
	for(int i=1;i<=n;++i)
	{
		cin>>a[i];
		what[a[i]]=i;
	}
	ll g=a[1],pos=1;
	for(int i=2;i<=n;++i)
		if(a[i]<g)
			g=a[i],pos=i;
	sort(a+1,a+n+1);
	reverse(a+1,a+n+1);
	solve();
	if(dfs(pos,pos,0)!=g)
	{
		cout<<-1<<endl;
		return 0;
	}
	for(int i=1;i<=n;++i)
		if(ans[i])
			cout<<i<<" "<<ans[i]<<endl;
	return 0;
}
