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
//I love Nanami Chiaki
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pii pair<int,int>
#define pli pair<ll,int>
#define mp make_pair
#define fst first
#define snd second
const int maxn=1e5+7;
int n;
ll d[maxn];
pli sx[maxn];
int sz[maxn];
ll dis[maxn];
int findit(ll t){
	if (t<0) return -1;
	int l=0,r=n-1,mid;
	while (l<=r){
		mid=(l+r)>>1;
		if (sx[mid].fst==t) return mid;
		if (sx[mid].fst<t) l=mid+1;
		else r=mid-1;
	}
	return -1;
}
void badend(){
	puts("-1");
	exit(0);
}
vector<pii> ans;
int main(){
	scanf("%d",&n);
	for (int i=0;i<n;i++){
		scanf("%lld",d+i);
		sx[i]=mp(d[i],i);
		sz[i]=1;dis[i]=0;
	}
	sort(sx,sx+n);
	for (int i=n-1;i>0;i--){
		int v=sx[i].snd;
		ll w=sx[i].fst;
		int u=findit(w+1ll*sz[v]*2-n);
		if (u==-1) badend();
		u=sx[u].snd;
		sz[u]+=sz[v];
		ans.push_back(mp(u+1,v+1));
		dis[u]+=dis[v]+sz[v];
	}
	if (dis[sx[0].snd]!=sx[0].fst) badend();
	for (int i=0;i<n-1;i++){
		printf("%d %d\n",ans[i].fst,ans[i].snd);
	}
	return 0;
}