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
#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define pb push_back
using namespace std;
using ll = long long;
using vi = vector<ll>;
const int maxn = 1<<19, mod = 1e9 + 7;
int n;
vi g[maxn];
ll t = 0;
void dfs(int v, int p = -1, int h = 0) {
	t += h;
	for(auto &i : g[v]) if(i != p) dfs(i, v, h + 1);
}
int main() {
	cin.tie(0)->sync_with_stdio(0);
	cin >> n;
	vi a(n), sz(n+1, 1);
	map<ll, int> id;
	for(int i = 0; i < n; i++) {
		cin >> a[i];
		id[a[i]] = i+1;
	}
	sort(rall(a));
	vector<pair<int, int>> edges;
	for(auto x : a) {
		int i = id[x];
		if(sz[i] == n) continue;
		ll nd = x + 2*sz[i] - n;
		if(nd < x && id.count(nd)) {
			edges.pb({i, id[nd]});
			sz[id[nd]] += sz[i];
		}
	}
	for(auto i : edges) {
		int x = i.first, y = i.second;
		g[x].pb(y);
		g[y].pb(x);
	}
	dfs(id[a.back()]);
	if(edges.size() != n-1 || t != a.back()) return cout << -1, 0;
	for(auto i : edges) cout << i.first << " " << i.second << '\n';
}
