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;
using ll = long long;
int main()
{
    int n;
    cin >> n;
    vector<ll> d(n);
    using P = pair<ll, ll>;
    vector<P> p(n);
    for(int i = 0; i < n; i++)
    {
        cin >> d[i];
        p[i] = P(d[i], i+1);
    }
    sort(d.begin(), d.end());
    sort(p.begin(), p.end());

    vector<int> num(n);
  vector<P> ans;
    vector<ll> dis(n);
    for(int i = n-1; i >= 1; i--)
    {
        num[i]++;
        ll cand = d[i] - n + 2 * num[i];
        int x = lower_bound(d.begin(), d.end(), cand) - d.begin();
        if(x >= i || d[x] != cand)
        {
            cout << -1 << endl;
            return 0;
        }
        num[x] += num[i];
        dis[x] = dis[x] + dis[i] + num[i];
        ans.push_back(P(x, i));
    }
    if(dis[0] != d[0])
    {
      cout << -1 << endl;
      return 0;
    }
    for(int i = 0; i < ans.size(); i++)
    {
        cout << p[ans[i].first].second << " " << p[ans[i].second].second << endl;
    }
}
