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>
using namespace std;
#define ll long long
#define mp make_pair
#define N 100100

ll n,cnt,dis;
ll d[N];
set<pair<ll,ll> > S;
ll sz[N],u[N],v[N];
set<pair<ll,ll> >::iterator it,it2;

int main()
{
	scanf("%lld",&n);
	for (ll i=1;i<=n;i++) {
		scanf("%lld",&d[i]);
		S.insert(mp(d[i],i));
		sz[i]=1;
	}
	
	for (ll i=1;i<n;i++)
	{
		it=S.end(); it--;
		ll now=(*it).second,num=sz[now];
		S.erase(it);
		if (n-2*num<=0) return puts("-1"),0;
		it2=S.lower_bound(mp(d[now]-(n-2*num),0));
		if (it2==S.end() || (*it2).first!=d[now]-(n-2*num)) return puts("-1"),0;
		u[++cnt]=now; v[cnt]=(*it2).second;
		sz[(*it2).second]+=num;
		dis+=2*num*(n-num);
	}
	for (int i=1;i<=n;i++) dis-=d[i];
	if (dis!=0) return puts("-1"),0;
	for (ll i=1;i<=cnt;i++) printf("%lld %lld\n",u[i],v[i]);
	return 0;
}