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>
#define pir pair<lli,lli>
#define mpa make_pair
#define fi first
#define se second
typedef long long lli;
using namespace std;
struct Edge{
	lli to,nxt;
}e[400010];
lli n,cnt,sum;
lli d[100010],h[100010],siz[100010],dis[100010];
map <lli,lli> id;
vector <pir> ans;
void add(lli u,lli v)
{
	e[++cnt].to=v;
	e[cnt].nxt=h[u];
	h[u]=cnt;
}
void dfs(lli u)
{
	for(lli i=h[u];i;i=e[i].nxt){
		lli v=e[i].to;
		dis[v]=dis[u]+1;
		dfs(v);
	}
}
int main()
{
	scanf("%lld",&n);
	for(lli i=1;i<=n;i++){
		scanf("%lld",&d[i]);
		id[d[i]]=i;
		siz[i]=1;
	}
	sort(d+1,d+n+1);
	for(lli i=n;i>1;i--){
		lli u=id[d[i]],tmp=n-2*siz[u];
		if(tmp<=0||id.count(d[i]-tmp)==0){
			puts("-1");
			return 0;
		}
		lli f=id[d[i]-tmp];
		add(f,u);
		ans.push_back(mpa(f,u));
		siz[f]+=siz[u];
	}
	dfs(id[d[1]]);
	for(lli i=1;i<=n;i++) sum+=dis[i];
	if(sum!=d[1]){
		puts("-1");
		return 0;
	}
	for(lli i=0;i<ans.size();i++)
		printf("%lld %lld\n",ans[i].fi,ans[i].se);
}