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>
using namespace std;

const int N = 1e5 + 5;
int n;
long long d[N];
priority_queue <pair <long long, int> > pq;
int siz[N];
int fa[N];
map <long long, set <int> > mp;
vector <int> g[N];

#define NO() return puts("-1"), 0

long long dfs(int u) {
	long long ans = 0;
	for (int i = 0; i < g[u].size(); ++ i)
		ans += dfs(g[u][i]) + siz[g[u][i]];
	return ans;
}

int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; ++ i)
		scanf("%lld", &d[i]), mp[d[i]].insert(i),
		siz[i] = 1, pq.push(make_pair(d[i], i));
	int rt;
	while (!pq.empty()) {
		int u = pq.top().second;
		pq.pop(); mp[d[u]].erase(u);
		if (!pq.empty()) {
			long long fa_d = (d[u] - (n - 2 * siz[u]));
//			printf("%d %lld %lld\n", u + 1, d[u], fa_d);
			if (!mp[fa_d].size()) NO();
			int v = *mp[fa_d].begin();
			fa[u] = v;
			g[v].push_back(u);
			siz[v] += siz[u];
		} else fa[u] = -1, rt = u;
	}
	if (dfs(rt) != d[rt]) NO();
	for (int i = 0; i < n; ++ i)
		if (~fa[i]) printf("%d %d\n", fa[i] + 1, i + 1);
	return 0;
}