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=100005;
int n,sz[Maxn],id[Maxn];
ll d[Maxn],sum;
vector<pair<int,int>>edge;
vector<int>son[Maxn];
void dfs1(int x,int dis=0){
    sum+=dis;
    for(int y:son[x])
        dfs1(y,dis+1);
}
void dfs2(int x,ll sum){
    if(sum!=d[x]){
        cout<<"-1\n";
        exit(0);
    }
    for(int y:son[x])
        dfs2(y,sum+n-2*sz[y]);
}
int main(){
    scanf("%d",&n);
    map<ll,int>Map;
    for(int i=1;i<=n;i++)scanf("%lld",&d[i]),Map[d[i]]=i;
    for(int i=1;i<=n;i++)id[i]=i;
    sort(id+1,id+n+1,[](int x,int y){return d[x]>d[y];});
    for(int i=1;i<n;i++){
        int x=id[i];
        sz[x]+=1;
        ll v=d[x]-n+2*sz[x];
        int y=Map[v];
        if(!y){
            cout<<"-1\n";
            return 0;
        }
        sz[y]+=sz[x];
        son[y].push_back(x);
        edge.emplace_back(x,y);
    }
    dfs1(id[n]);
    dfs2(id[n],sum);
    for(auto i:edge)
        cout<<i.first<<" "<<i.second<<"\n";
    return 0;
}