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;

#define int long long
const double PI = 3.14159265358979323846;
typedef vector<int> vint;
typedef pair<int, int> pint;
int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};

int N;
int D[110000], n[110000], d[110000];
unordered_map<int, int> mp;
vint ans;
signed main() {
    cin >> N;
    for (int i = 1; i <= N; i++) {cin >> D[i]; n[i] = 1; mp[D[i]] = i;};
    sort(D + 1, D + N + 1); reverse(D + 1, D + N + 1);
    for (int i = 1; i <= N - 1; i++) {
        int s = mp[D[i]];
        int t = mp[D[i] - N + 2 * n[s]];
        n[t] += n[s];
        d[t] += d[s] + n[s];
        ans.push_back(s); ans.push_back(t);
    }
    if (n[mp[D[N]]] != N || d[mp[D[N]]] != D[N]) {
        cout << -1 << endl;
        return 0;
    }
    for (int i = 0; i < 2 * (N - 1); i += 2) {
        cout << ans[i] << " " << ans[i + 1] << endl;
    }

}
