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
#define re register
#define ull unsigned ll
using namespace std;
inline ll read(){
	ll s=0,t=1;
	char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')t=-1;ch=getchar();}
	while(ch>='0'&&ch<='9')s=(s<<3)+(s<<1)+(ch^48),ch=getchar();
	return s*t;
}
#define f1 first
#define f2 second
#define mp make_pair
const int N=1e5+5;
int n,sz[N],f[N];ll Sum;
vector<int> G[N];
pair<ll,int> p[N];
void DFS(int x,ll d){Sum+=d;for(auto y:G[x])DFS(y,d+1);}
int main(){
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	n=read();
	for(int i=1;i<=n;i++)sz[i]=1,p[i]=mp(read(),i);
	sort(p+1,p+n+1);
	for(int i=n;i>1;i--){
		int x=p[i].f2;
		ll v=p[i].f1+2*sz[x]-n;
		int j=lower_bound(p+1,p+i,mp(v,0))-p;
		if(j>n)return puts("-1"),0;
		G[f[x]=p[j].f2].push_back(x),sz[f[x]]+=sz[x];
	}DFS(p[1].f2,0);
	if(Sum^p[1].f1)return puts("-1"),0;
	for(int i=1;i<=n;i++)if(f[i])printf("%d %d\n",f[i],i);
	return 0;
}
