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;

int main() {
#ifdef LOCAL
  freopen("input.txt", "r", stdin);
#endif
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<pair<long long, int>> d(n);
  for (int i = 0; i < n; ++i) {
    cin >> d[i].first;
    d[i].second = i;
  }
  sort(d.begin(), d.end());
  vector<pair<int, int>> ans;
  vector<long long> sum(n);
  vector<int> sz(n, 1);
  for (int i = n - 1; i; --i) {
    long long v = d[i].first - n + 2 * sz[i];
    int p = lower_bound(d.begin(), d.end(), make_pair(v, 0)) - d.begin();
    if (p == n || d[p].first != v) {
      cout << -1 << "\n";
      return 0;
    }
    sum[p] += sum[i] + sz[i];
    sz[p] += sz[i];
    ans.emplace_back(d[i].second, d[p].second);
  }
  if (sum[0] != d[0].first) {
    cout << -1 << "\n";
    return 0;
  }
  for (auto p : ans) {
    cout << p.first + 1 << " " << p.second + 1 << "\n";
  }
  return 0;
}
