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 Rust (1.42.0) (judge verdict: Accepted)

use proconio::input;

fn err() -> ! {
    println!("{}", -1);
    std::process::exit(0);
}

fn main() {
    input!(n: usize, d: [u64; n]);
    let mut ord: Vec<_> = (0..n).collect();
    ord.sort_by_key(|&i| std::cmp::Reverse(d[i]));
    let mut vid = vec![0; n];
    (0..n).for_each(|i| vid[ord[i]] = i);
    let mut d_sorted_reverse = d.clone();
    d_sorted_reverse.sort_by_key(|&x| std::cmp::Reverse(x));

    let mut g = vec![vec![]; n];
    let mut ans = vec![];
    let mut size = vec![1; n];
    for (si, &i) in ord.iter().enumerate().take(n - 1) {
        if n as u64 <= size[i] * 2 {
            err();
        }
        let di = d[i];
        let dj = di - n as u64 + 2 * size[i];
        let sj = d_sorted_reverse
            .binary_search_by_key(&std::cmp::Reverse(dj), |&x| std::cmp::Reverse(x))
            .unwrap_or_else(|_| err());
        assert!(si < sj);
        let j = ord[sj];
        size[j] += size[i];
        ans.push((i, j));
        g[j].push(i);
    }

    let root = ord[n - 1];
    fn dfs(x: usize, g: &Vec<Vec<usize>>, size: &Vec<u64>) -> u64 {
        let mut ans = 0;
        for &y in &g[x] {
            ans += dfs(y, &g, &size) + size[y];
        }
        ans
    }
    if dfs(root, &g, &size) != d[root] {
        err();
    }
    for (u, v) in ans {
        println!("{} {}", u + 1, v + 1);
    }
}