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
#include <cstdio>
#include <algorithm>
#include <vector>
#define file(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)

const int N(1e5 + 10);
struct edge { int x, y; } e[N];
struct node { long long d; int id; } a[N];
inline int operator < (const node &l, const node &r)
	{ return l.d < r.d; }
int n, size[N], cnt;

int main()
{
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
		scanf("%lld", &a[i].d), a[i].id = i, size[i] = 1;
	std::sort(a + 1, a + n + 1);
	long long dis = 0, d;
	for (int i = n, p; i > 1; i--)
	{
		d = a[i].d - n + (size[i] << 1), p = std::lower_bound(a + 1, a + n + 1, (node) {d, 0}) - a;
		if (a[p].d != d) return puts("-1"), 0;
		e[++cnt] = (edge) {a[i].id, a[p].id}, size[p] += size[i], dis += size[i];
	}
	if (dis != a[1].d) puts("-1");
	else for (int i = 1; i < n; i++) printf("%d %d\n", e[i].x, e[i].y);
	return 0;
}
