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<bits/stdc++.h>
using namespace std;
#define ll long long

map<ll,int>id;
int n;
ll d[101000];

#define pli pair<ll,int>
#define fi first
#define se second
priority_queue<pli>q;

int fa[101000], sz[101000]; ll sum[101000];

int main(){
    cin>>n;
    for (int i=1;i<=n;++i){
        cin>>d[i]; id[d[i]]=i;
    }
    sort(d+1,d+n+1,greater<ll>());
    for (int i=1;i<=n;++i){
        sz[i]=1;
        while (q.size()&&q.top().fi==d[i]){
            fa[q.top().se]=i;
            sz[i]+=sz[q.top().se];
            sum[i]+=sum[q.top().se]+sz[q.top().se];
            q.pop();
        }
        q.push({d[i]-(n-sz[i]*2),i});
    }
    if (q.size()!=1||sum[n]!=d[n]){
        puts("-1");
        return 0;
    }
    for (int i=1;i<n;++i)
        printf("%d %d\n",id[d[i]],id[d[fa[i]]]);
}