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<stdio.h>
#include<algorithm>
#include<iostream>
#include<map>
#include<vector>
#define ll long long
#define pii pair<int,int>
#define MP make_pair
using namespace std;
ll n;
map<ll,int>rev;
struct node{ll d,siz,val;int id;}a[100005];
bool cmp(node x,node y){return x.d>y.d;}
int back[100005];
vector<pii>ans;
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%lld",&a[i].d);
        a[i].siz=1;a[i].id=i;
        rev[a[i].d]=i;
    }
    sort(a+1,a+1+n,cmp);
    for(int i=1;i<=n;i++) back[a[i].id]=i;
    for(int i=1;i<n;i++){
        if(n-2*a[i].siz<=0){puts("-1");return 0;}
        ll dd=a[i].d+2*a[i].siz-n;
        if(!rev[dd]){puts("-1");return 0;}
        int fa=rev[dd];
        ans.push_back(MP(min(fa,a[i].id),max(fa,a[i].id)));
        a[back[fa]].siz+=a[i].siz;
        a[back[fa]].val+=a[i].val+a[i].siz;
    }
    if(a[n].d!=a[n].val) return puts("-1"),0;
    sort(ans.begin(),ans.end());
    for(int i=0;i<n-1;i++) printf("%d %d\n",ans[i].first,ans[i].second);
    return 0;
}
