Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void err() {
  cout << -1 << endl;
  exit(0);
}
void dfs (int crr, int dpt, vector<vector<int>>& grh, long long& cnt) {
  cnt += dpt;
  for (int nxt : grh[crr]) {
    dfs(nxt, dpt + 1, grh, cnt);
  }
}
int main() {
  int n;
  cin >> n;
  vector<long long> d(n);
  map<long long, int> mp;
  for (int i = 0; i < n; i++) {
    cin >> d[i];
    mp[d[i]] = i;
  }
  sort(d.begin(), d.end(), greater<long long>());
  vector<long long> sz(n, 1);
  vector<vector<int>> grh(n);
  for (auto it = d.begin(); next(it) != d.end(); it++) {
    auto i = mp[*it];
    long long ptv = *it - n + 2 * sz[i];
    auto jt = lower_bound(d.rbegin(), d.rend(), ptv);
    if (jt == d.rend() || *jt != ptv) err();
    auto p = mp[*jt];
    grh[p].push_back(i);
    sz[p] += sz[i];
  }
  long long cnt = 0;
  int rt = mp[d[n - 1]];
  dfs(rt, 0, grh, cnt);
  if (d[n - 1] != cnt) err();
  for (int crr = 0; crr < n; crr++) {
    for (int nxt : grh[crr]) {
      cout << crr + 1 << ' ' << nxt + 1 << endl;
    }
  }
  return 0;
}