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;
const int MX=1e5+5;
pair<long long,int> a[MX];
map<long long,int> mp;
long long n;
int p[MX];
int visit[MX];
long long s[MX];
long long dp[MX];
int main(){
    cin.tie(0);
    cout.tie(0);
    ios_base::sync_with_stdio(0);
    cin>>n;
    int i;
    for(i=1 ; i<=n ; i++){
        cin>>a[i].first;
      ///  a[i].first++;
        a[i].second=i;
        mp[a[i].first]=i;
    }
    sort(a+1,a+n+1);
    for(i=n ; i>=2 ; i--){
      ///  cout<<i<<endl;
        s[a[i].second]++;

        int k=mp[a[i].first+2*s[a[i].second]-n];
        if(k==0 || k==a[i].second || a[i].first+2*s[a[i].second]-n>=a[i].first){
            cout<<-1;
            return 0;
        }
        s[k]+=s[a[i].second];
        dp[k]+=dp[a[i].second]+s[a[i].second];
        p[a[i].second]=k;
    }
    if(dp[a[1].second]!=a[1].first){
        cout<<-1;
        return 0;
    }
    for(i=2 ; i<=n ;i++){
        cout<<a[i].second<<" "<<p[a[i].second]<<"\n";
    }
    return 0;
}
