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 n;
long long d[100001];
int x[100001];
int size[100001];
long long g[100001];
vector<pair<int, int> > edge;
map<long long, int>f;
bool cmp(int x, int y) {
    return d[x] > d[y];
}
int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n ; i ++)
        scanf("%lld", &d[i]), x[i] = i, size[i] = 1, f[d[i]] = i;
    sort(x + 1, x + n + 1, cmp);
    bool sig = 0;
    for (int i = 1; i <= n ; i ++) {
        long long dd = d[x[i]] +size[x[i]] - n + size[x[i]];
        if (dd != d[x[i]]&&f[dd] && dd < d[x[i]]) size[f[dd]] += size[x[i]], edge.push_back(make_pair(x[i], f[dd])), g[f[dd]] += g[x[i]] + size[x[i]];
    }
    if (size[x[n]] == n && edge.size() == n - 1 &&g[x[n]] == d[x[n]]){
        for (auto u : edge) printf("%d %d\n", u.first, u.second);
    } else printf("-1");
    return 0;
}