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<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#define rep(i,l,r) for (int i=(l); i<=(r); i++)
typedef long long ll;
using namespace std;

const int N=100010;
pair<long long,int> p[N];
int n,s[N];
ll S[N];
vector< pair<int,int> > ans;

int main(){
	scanf("%d",&n);
	rep(i,1,n) scanf("%lld",&p[i].first),p[i].second=i,s[i]=1;
	sort(p+1,p+n+1);
	for(int i=n; i>1; --i){
		if (n-2*s[i]<0) { puts("-1"); return 0; }
		int x=lower_bound(p+1,p+n+1,make_pair(p[i].first-n+2*s[i],0))-p;
		if(p[x].first!=p[i].first-n+2*s[i]) { puts("-1"); return 0; }
		s[x]+=s[i]; S[x]+=S[i]+s[i]; ans.push_back(make_pair(p[x].second,p[i].second));
	}
	if (S[1]!=p[1].first) { puts("-1"); return 0; }
	for(int i=0; i<(int)ans.size(); ++i) printf("%d %d\n",ans[i].first,ans[i].second);
	return 0;
}