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 <cctype>
#include <algorithm>
#include <utility>
#define MAXN 100005
using namespace std;
typedef long long ll;
typedef pair<ll,int> pi;
ll read()
{
	ll ans=0;
	char c=getchar();
	while (!isdigit(c)) c=getchar();
	while (isdigit(c)) ans=(ans<<3)+(ans<<1)+(c^48),c=getchar();
	return ans;
}
pi p[MAXN];
int siz[MAXN],fa[MAXN];
ll d[MAXN];
int main()
{
	int n=read();
	for (int i=1;i<=n;i++) p[i]=make_pair(read(),i),siz[i]=1;
	sort(p+1,p+n+1);
	ll sum=0;
	for (int i=n;i>1;i--)
	{
		ll t=p[i].first-n+2*siz[p[i].second];
		int k=lower_bound(p+1,p+n+1,make_pair(t,0))-p;
		if (p[k].first!=t) return puts("-1"),0;
		fa[p[i].second]=p[k].second,siz[p[k].second]+=siz[p[i].second];
		sum+=siz[p[i].second];
	}
	if (sum!=p[1].first) return puts("-1"),0;
	for (int i=1;i<=n;i++) if (fa[i]) printf("%d %d\n",fa[i],i);
	return 0;
}