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<cstring>
#include<algorithm>
#include<map>
#include<vector>
using namespace std;
struct node
{
	long long d;
	int id;
	friend bool operator < (node a,node b){return a.d>b.d;}
}p[100010];
int n,siz[100010],ans[100010][2];
map<long long,int>mp;
vector<int>E[100010];
long long sum;
void dfs(int v,int dep)
{
	sum+=dep;
	for(auto u:E[v])dfs(u,dep+1);
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;++i)scanf("%lld",&p[i].d),p[i].id=i;
	sort(p+1,p+1+n);
	for(int i=1;i<=n;++i)mp[p[i].d]=p[i].id,siz[i]=1;
	for(int i=1;i<n;++i)
	{
	//	printf("%lld %d\n",p[i].d,n-2*siz[p[i].id]);
		mp.erase(p[i].d);
		if(!mp.count(p[i].d-n+2*siz[p[i].id]))return printf("-1"),0;
		ans[i][0]=p[i].id,ans[i][1]=mp[p[i].d-n+2*siz[p[i].id]];
		siz[ans[i][1]]+=siz[ans[i][0]];
		E[ans[i][1]].emplace_back(ans[i][0]);
	}
	dfs(p[n].id,0);
	if(sum!=p[n].d)return printf("-1"),0;
	for(int i=1;i<n;++i)printf("%d %d\n",ans[i][0],ans[i][1]);
	return 0;
}