Problem: 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

Solution (in CPP):

Solution:
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long ll;
ll n, b[100005], fa[100005], length[100005], f[100005];
bool islead[100005];
struct lala
{
	ll v, id;
	friend bool operator < (lala a, lala b)
	{
		return a.v > b.v;
	}
}a[100005];

ll ef(ll ls, ll rs, ll val)
{
	if (ls == rs) return ls;
	ll mid = (ls + rs) / 2;
	if (a[mid].v <= val) return ef(ls, mid, val);
	else return ef(mid + 1, rs, val);
}

int main()
{
	cin >> n;
	for (ll i = 1; i <= n; i++) 
	{
		cin >> a[i].v;
		a[i].id = i;
		b[i] = a[i].v;
	}
	sort(a + 1, a + n + 1);
	for (ll i = 1; i <= n; i++)
	{
		length[i] = 1;
	}
	for (ll x = 1; x <= n; x++)
	{
		ll i = ef(1, n, a[x].v - n + length[x] * 2);
		if (a[i].v == a[x].v - n + length[x] * 2 && a[x].v > a[i].v)
		{
			fa[x] = i;
			length[i] += length[x];
			f[i] += f[x] + length[x];
		}
	}
	if (f[n] != a[n].v)
	{
		cout << "-1" << endl;
		return 0;
	}
	for (ll i = 1; i < n; i++)
	{
		cout << a[i].id << " " << a[fa[i]].id << endl;
	}
	return 0;
}
/*
7 10 15 13 18 11 14 19
*/