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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 100233;
int n, size[N], fa[N], dep[N];
pair<int, ll> p[N];
map<ll, int> q;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &p[i].second);
    p[i].first = i;
    q[p[i].second] = i;
    size[i] = 1;
  }
  sort(p + 1, p + 1 + n, [&](pair<int, ll> a, pair<int, ll> b){return a.second < b.second;});
  for (int i = n; i >= 2; i--) {
    int x = p[i].first;
    ll t = p[i].second + 2 * size[x] - n;
    if (t >= p[i].second) return puts("-1"), 0;
    int y = q[t];
    if (!y) return puts("-1"), 0;
    size[y] += size[x];
    fa[x] = y;
  }
  ll t = 0;
  for (int i = 2; i <= n; i++) {
    int x = p[i].first;
    dep[x] = dep[fa[x]] + 1;
    t += dep[x];
  }
  if (t != p[1].second) return puts("-1"), 0;
  for (int i = 2; i <= n; i++) {
    int x = p[i].first;
    printf("%d %d\n", x, fa[x]);
  }
}
