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
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1000005
#define inf 1e9
#define eps 1e-10
using namespace std;

inline ll read()
{
	ll x=0,w=1; char c=getchar();
	while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
	while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
	return w==1?x:-x;
}

ll n,d[maxn],siz[maxn],f,dis;
map <ll,int> p;
vector <int> a1,a2;

inline bool cmp(ll a,ll b){return a>b;}

int main()
{
	n=read(); f=0;
	for(int i=1;i<=n;i++) d[i]=read(),p[d[i]]=i;
	sort(d+1,d+n+1,cmp);
	for(int i=1;i<=n;i++)
	{
		siz[p[d[i]]]++;
		ll dfa=d[i]-n+2*siz[p[d[i]]];
		if(dfa>d[i]) continue;
		siz[p[dfa]]+=siz[p[d[i]]]; dis+=siz[p[d[i]]];
		//cout<<d[i]<<" "<<p[d[i]]<<" "<<siz[p[d[i]]]<<" "<<dfa<<endl;
		if(p[dfa]==0) {f=1; break;}
		a1.push_back(p[d[i]]),a2.push_back(p[dfa]);
	}
	if(f==1||dis!=d[n]) {puts("-1"); return 0;}
	for(int i=0;i<a1.size();i++) printf("%d %d\n",a1[i],a2[i]);
	return 0;
}