Problem:
Score : 900 points

Problem Statement

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.

Sample Input 1

7
10
15
13
18
11
14
19

Sample Output 1

1 2
1 3
1 5
3 4
5 6
6 7

The tree shown below satisfies the conditions.

Sample Input 2

2
1
2

Sample Output 2

-1

Sample Input 3

15
57
62
47
45
42
74
90
75
54
50
66
63
77
87
51

Sample Output 3

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

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>

using namespace std;
typedef long long int ll;

const int MAXN = 1e5 + 5;
int n, par[MAXN], d[MAXN];
ll sz[MAXN];
vector <pair<ll,int> > v;
map <ll,int> mp;

int main()
{
	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	cin >> n;
	for(int i = 1; i <= n; i++){
		cin >> d[i];
		mp[d[i]] = i; sz[i] = 1;
		v.push_back({d[i], i});
	}
	sort(v.begin(), v.end());
	for(int i = n - 1; i > 0; i--){
		int ind = v[i].second, p;
		ll dd = v[i].first, dpar = dd - (ll)n + (2ll * sz[ind]);
		p = mp[dpar];
		if(p == 0 || p == ind){
			cout << -1 << endl;
			return 0;
		}
		par[ind] = p;
		sz[p] += sz[ind];
	}
	for(int i = 1; i < n; i++){
		int ind = v[i].second;
		cout << ind << " " << par[ind] << '\n';
	}
	return 0; 
}