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 <numeric>
#include <map>
#define fail return puts("-1"), 0

const int N = 1e5 + 5, M = 2e5 + 5;

int n, tot, lnk[N], ter[M], nxt[M], sz[N], fa[N];
long long dis[N], d[N];
std::map<long long, int> mp;

void add(int u, int v) {
	ter[++tot] = v, nxt[tot] = lnk[u], lnk[u] = tot;
}
void dfs(int u, int p) {
	for (int i = lnk[u]; i; i = nxt[i]) {
		int v = ter[i];
		if (v == p) continue;
		dis[v] = dis[u] + 1;
		dfs(v, u);
	}
}
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%lld", &d[i]);
		mp[d[i]] = i, sz[i] = 1;
	}
	int cnt = 0;
	for (auto it = mp.rbegin(); it != mp.rend(); it++) {
		int v = it->second, u = mp[it->first + 2 * sz[v] - n];
		if (!u || u == v) fail;
		add(u, v), add(v, u);
		fa[v] = u, sz[u] += sz[v];
		if (++cnt == n - 1) break;
	}
	int rt = mp.begin()->second;
	dfs(rt, 0);
	if (std::accumulate(dis + 1, dis + n + 1, 0LL) != d[rt]) fail;
	for (int i = 1; i <= n; i++) {
		if (fa[i]) printf("%d %d\n", fa[i], i);
	}
	return 0;
}