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 int N = 1e5 + 10;
long long rn, n, d[N], s[N];
vector<int> gr[N];
map<long long, int> mp;
void dfs(int v, int h = 0) {
	for (auto u : gr[v])
		dfs(u, h + 1);
	rn += h;
}
int32_t main() {
	ios_base::sync_with_stdio (false), cin.tie(0), cout.tie(0);
	cin >> n;
	for (int i = 0; i < n; i++)
		cin >> d[i], mp[d[i]] = i;
	sort(d, d + n, greater<int> ());
	fill(s, s + n + 1, 1);
	for (int i = 0; i < n - 1; i++) {
		int v = mp[d[i]], valpar = d[i] - n + 2 * s[v];
		if (!mp.count(valpar)) {
			cout << -1;
			return 0;
		}
		int par = mp[valpar];
		if (par == v) {
			cout << -1;
			return 0;
		}
		s[par] += s[v], gr[par].push_back(v);
	}
	dfs(mp[d[n - 1]]);
	if (rn != d[n - 1]) {
		cout << -1;
		return 0;
	}
	for (int i = 0; i < n; i++) {
		for (auto u : gr[i])
			cout << i + 1 << " " << u + 1 << "\n";
	}
	return 0;
}
