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;
typedef long long ll;
typedef pair <ll, int> pli;

const int N = 1e5 + 10;
int n, cnt[N], par[N], mark[N];
ll sum[N];
pli ar[N];
unordered_map <ll, int> mp;
vector <int> ch[N];

void dfs(int v) {
	mark[v] = 1;
	for (int i : ch[v]) {
		dfs(i);
		sum[v] += sum[i] + cnt[i];
	}
}

int32_t main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	fill(cnt, cnt + N, 1);
	
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> ar[i].first;
		ar[i].second = i;
		mp[ar[i].first] = i;
	}
	sort(ar, ar + n);
	
	for (int i = n - 1; i > 0; i--) {
		int v = ar[i].second;
		ll nd = ar[i].first - n + (ll)2 * cnt[v];
		if (mp.find(nd) == mp.end() || 2 * cnt[v] > n)
			return cout << -1, 0;
		
		int p = mp[nd];
		ch[p].push_back(v);
		par[v] = p;
		cnt[p] += cnt[v];
	}
	
	dfs(ar[0].second);
	if (sum[ar[0].second] != ar[0].first)
		return cout << -1, 0;
	
	for (int i = 0; i < n; i++)
		if (!mark[i])
			return cout << -1, 0;
	
	for (int i = 0; i < n; i++)
		if (i != ar[0].second)
			cout << i + 1 << " " << par[i] + 1 << '\n';
}