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 <bits/stdc++.h>

template <class T>
inline void read(T &res)
{
	res = 0; bool bo = 0; char c;
	while (((c = getchar()) < '0' || c > '9') && c != '-');
	if (c == '-') bo = 1; else res = c - 48;
	while ((c = getchar()) >= '0' && c <= '9')
		res = (res << 3) + (res << 1) + (c - 48);
	if (bo) res = ~res + 1;
}

typedef long long ll;

const int N = 1e5 + 5;

int n, p[N], fa[N], sze[N], dis[N];
ll d[N], dsum;
std::map<ll, int> occ;

inline bool comp(int a, int b) {return d[a] < d[b];}

int main()
{
	read(n);
	for (int i = 1; i <= n; i++) read(d[i]), p[i] = i, sze[i] = 1;
	std::sort(p + 1, p + n + 1, comp);
	for (int i = 1; i <= n; i++) occ[d[i]] = i;
	for (int i = n; i >= 2; i--)
	{
		int u = p[i]; occ[d[u]] = 0;
		int v = occ[d[u] - n + (sze[u] << 1)];
		if (!v) return puts("-1"), 0;
		fa[u] = v; sze[v] += sze[u];
	}
	for (int i = 2; i <= n; i++) dis[p[i]] = dis[fa[p[i]]] + 1;
	for (int i = 1; i <= n; i++) dsum += dis[i];
	if (d[p[1]] != dsum) return puts("-1"), 0;
	for (int i = 2; i <= n; i++) printf("%d %d\n", fa[p[i]], p[i]);
	return 0;
}