Problem: 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

Solution (in CPP):

Solution:
#include<cstdio>
#include<map>
#include<algorithm>
using namespace std;
#define int long long
int n,sz[100100],u[100100],v[100100],head[100100],cnt,D;
struct Edge{
	int to,nxt;
}edge[200100];
struct node{
	int dis,x;
	bool operator<(node b)const{
		return dis>b.dis;
	}
}a[100100];
map<int,int>mp;
void addedge(int x,int y){
	edge[++cnt].to=y;
	edge[cnt].nxt=head[x];
	head[x]=cnt;
}
int dfs(int x,int fa,int now){
	D+=now;
	for(int i=head[x];i;i=edge[i].nxt){
		int upup=edge[i].to;
		if(upup==fa)continue;
		dfs(upup,x,now+1);
	}
}
signed main(){
	scanf("%lld",&n);
	for(int i=1;i<=n;i++){
		scanf("%lld",&a[i].dis);
		a[i].x=i;
		mp[a[i].dis]=i;
	}
	sort(a+1,a+n+1);
	for(int i=1;i<=n;i++)sz[i]=1;
	for(int i=1;i<n;i++){
		int dis=a[i].dis,x=a[i].x;
		int tmp=dis-abs(n-2*sz[x]);
		int fa=mp.find(tmp)->second;
		if(fa==x||!fa)return puts("-1"),0;
		sz[fa]+=sz[x];
		D+=sz[a[i].x];
		u[i]=x,v[i]=fa;
		addedge(x,fa);
		addedge(fa,x);
	}
	//dfs(a[n].x,0,0);
	if(D!=a[n].dis)return puts("-1"),0;
	for(int i=1;i<n;i++)
		printf("%lld %lld\n",u[i],v[i]);
}