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 <map>
#include <vector>
#include <algorithm>
#include <numeric>
#include <iostream>
using namespace std;
int main() {
    int n; cin >> n;
    long long d[n]; for (int i = 0; i < n; i++) cin >> d[i];
    int v[n];
    iota(v, v+n, 0);
    sort(v, v+n, [&d](int i, int j) { return d[i] > d[j]; });
    map<long long, int> vertex;
    for (int i = 0; i < n; i++) vertex[d[i]] = i;
    vector<pair<int, int> > ans;
    int sz[n];
    fill(sz, sz+n, 1);
    for (int p = 0; p < n-1; p++) {
        const int i = v[p];
        const long long dist = d[i] - n + 2 * sz[i];
        // cerr << i << " " << sz[i] << " " << d[i] << " " << dist << endl;
        if (d[i] == dist || vertex.find(dist) == vertex.end()) {
            cout << "-1\n";
            return 0;
        }
        int j = vertex[dist];
        sz[j] += sz[i];
        ans.push_back({j, i});
    }
    if (d[v[n-1]] == accumulate(sz, sz+n, 0LL) - sz[v[n-1]])
        for (auto [i, j] : ans) cout << i+1 << " " << j+1 << endl;
    else
        cout << "-1\n";
}