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;
void ensure(bool exp) {
    if (!exp) {
        cout << -1 << endl;
        exit(0);
    }
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    cin >> n;
    vector<long long> d(n);
    map<long long, int> ver;
    for (int i = 0; i < n; ++i) {
        cin >> d[i];
        ver[d[i]] = i;
    }
    vector<int> p(n);
    iota(p.begin(), p.end(), 0);
    sort(p.begin(), p.end(), [&](int i, int j) {
        return d[i] < d[j];
    });
    vector<int> sz(n, 1), parent(n, -1);
    vector<long long> f(n);
    for (int i = n - 1; i >= 1; --i) {
        int u = p[i];
        long long dp = d[u] - n + 2 * sz[u];
        ensure(dp < d[u]);
        ensure(ver.count(dp) > 0);
        parent[u] = ver[dp];
        sz[parent[u]] += sz[u];
        f[parent[u]] += f[u] + sz[u];
    }
    ensure(f[p[0]] == d[p[0]]);
    for (int i = 1; i < n; ++i)
        cout << parent[p[i]] + 1 << " " << p[i] + 1 << "\n";
    return 0;
}