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>

typedef long long LL;
const int MN = 100005;

int N, p[MN], siz[MN], ok;
LL dis[MN], sum[MN];
bool cmp(int i, int j) { return dis[i] > dis[j]; }
std::vector<int> G[MN];

void DFS(int u, LL d) {
	if (d != dis[u]) ok = 0;
	for (auto v : G[u]) DFS(v, d + N - 2 * siz[v]);
}

int main() {
	scanf("%d", &N);
	for (int i = 1; i <= N; ++i) scanf("%lld", &dis[i]), p[i] = i, siz[i] = 1;
	std::sort(p + 1, p + N + 1, cmp);
	for (int i = 1; i < N; ++i) {
		int u = p[i];
		if (2 * siz[u] >= N) return puts("-1"), 0;
		dis[0] = dis[u] - N + 2 * siz[u];
		int pos = std::lower_bound(p + i + 1, p + N + 1, 0, cmp) - p;
		if (pos > N) return puts("-1"), 0;
		int v = p[pos];
		G[v].push_back(u);
		siz[v] += siz[u];
		sum[v] += sum[u] + siz[u];
	}
	ok = 1;
	DFS(p[N], sum[p[N]]);
	if (!ok) return puts("-1"), 0;
	for (int i = 1; i <= N; ++i)
		for (auto v : G[i]) printf("%d %d\n", i, v);
	return 0;
}