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 <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
#include <map>
#include <cstdio>
#define int long long
#define pb push_back
#define mp make_pair
#define INF 0x3f3f3f3f
#define Inf 1000000000000000000LL
#define F first
#define S second
using namespace std;
typedef pair<int,int>pii;
int n;
map<int,int>ret;
vector<pii>e;
signed main(){
	
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	
	cin>>n;
	for(int i=0LL,x;i<n;i++){
		cin>>x;
		ret[x]=i;
		e.pb(mp(x,i));
	}
	sort(e.rbegin(),e.rend());
	int x=e.back().F;
	vector<int>sz(n,1LL);
	vector<int>dp(n,0LL);
	vector<pii>ans;
	for(auto c:e){
		if(c.F!=x){
			if(sz[c.S]*2LL>=n)return puts("-1"),0;
			int p=c.F-(n-2LL*sz[c.S]);
			if(!ret.count(p))return puts("-1"),0;
			int pa=ret[p];
			dp[pa]+=dp[c.S]+sz[c.S];
			sz[pa]+=sz[c.S];
			ans.pb(mp(c.S+1LL,pa+1LL));
		}
	}
	if(dp[e.back().S]!=x)return puts("-1"),0;
	for(auto c:ans)cout<<c.F<<' '<<c.S<<endl;
	return 0;
}
