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;
typedef long long ll;
const int maxn=100010;
struct node{
	ll sum;int id;
	bool operator<(const node n)const{
		return sum>n.sum;
	}
}p[maxn];
int n,sz[maxn],fa[maxn];
int to[maxn],nxt[maxn],head[maxn],el;
ll sum[maxn],tot;
inline void add(int u,int v){
	to[++el]=v;nxt[el]=head[u];head[u]=el;
}
void dfs(int u,ll dis){
	tot+=dis;
	for(int i=head[u];i;i=nxt[i]) dfs(to[i],dis+1);
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%lld",&p[i].sum),p[i].id=i,sz[i]=1;
	sort(p+1,p+n+1);
	for(int i=1;i<=n;i++) sum[i]=p[i].sum;
	for(int i=1;i<n;i++){
		int f=lower_bound(sum+1,sum+n+1,sum[i]-n+2*sz[i],greater<ll>())-sum;
		if(sum[f]!=sum[i]-n+2*sz[i] || f==i) return puts("-1"),0;
		fa[p[i].id]=p[f].id;sz[f]+=sz[i];add(f,i);
	}
	dfs(n,0);
	if(tot!=sum[n]) return puts("-1"),0;
	for(int i=1;i<=n;i++) if(fa[i]) printf("%d %d\n",i,fa[i]);
}