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 pair<int64_t, int64_t> Pair;

void fail(){
    cout << -1 << endl;
    exit(0);
}

int main(){
    int i, j, k;
    int N;
    cin >> N;
    vector<Pair> D(N);

    for(i=0; i<N; i++){
        int64_t d;
        cin >> d;
        D.push_back({d, i});
    }
    sort(D.rbegin(), D.rend());

    map<int64_t, int64_t> mp;
    for(i=0; i<N; i++) mp[D[i].first] = i;

    vector<Pair> ans;
    vector<int64_t> dist_sub(N, 0);
    vector<int> num_sub(N, 1);

    for(i=0; i<N-1; i++){
        int64_t diff = N - 2*num_sub[i];
        int64_t pd = D[i].first - diff;
        if(mp.count(pd)){
            int p = mp[pd];
            ans.push_back({D[i].second+1, D[p].second+1});
            num_sub[p] += num_sub[i];
            dist_sub[p] += dist_sub[i] + num_sub[i];
        }else{
            fail();
        }
    }
    if(dist_sub[N-1] != D[N-1].first) fail();

    for(auto&& e : ans){
        cout << e.first << " " << e.second << endl;
    }
    return 0;
}