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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define eb emplace_back

using namespace std;
typedef long long ll;
typedef pair<int, int> ii;

const int maxn = 1e5 + 5;

int par[maxn], N, sub[maxn];
pair<ll, int> d[maxn];

signed main(void)
{
  ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
  if(fopen("A.INP", "r")){
    freopen("A.INP", "r", stdin);
    freopen("A.OUT", "w", stdout);
  }
  cin >> N;
  for(int i = 1; i <= N; ++i){
    cin >> d[i].fi;
    d[i].se = i;
    sub[i] = 1;
  }
  sort(d + 1, d + 1 + N);
  for(int i = N; i >= 2; --i){
    int j = lower_bound(d + 1, d + 1 + N, mp(d[i].fi - (N - 2 * sub[d[i].se]), -1)) - d;
    if(i == j) ++j;
    if(d[j].fi != d[i].fi - (N - 2 * sub[d[i].se])){
      cout << -1;
      return 0;
    }
    par[d[i].se] = d[j].se;
    sub[d[j].se] += sub[d[i].se];
  }
  for(int i = 1; i <= N; ++i){
    if(par[i]) cout << i << ' ' << par[i] << '\n';
  }
}