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;

static const int Maxn = 100005;

int n, id[Maxn];
int par[Maxn], sz[Maxn];
long long a[Maxn];
map<long long, int> mp;

int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", a + i);
    id[i] = i, sz[i] = 1;
    mp[a[i]] = i;
  }
  sort(id + 1, id + n + 1, [&](const int &x, const int &y)->bool {
    return a[x] > a[y];
  });
  for (int i = 1; i < n; ++i) {
    int now = id[i];
    par[now] = mp[a[now] - n + 2 * sz[now]];
    if (!par[now]) return puts("-1") & 0;
    sz[par[now]] += sz[now];
  }
  long long sum = 0;
  for (int i = 1; i < n; ++i) sum += sz[id[i]];
  if (sum != a[id[n]]) return puts("-1") & 0;
  for (int i = 1; i < n; ++i)
    printf("%d %d\n", id[i], par[id[i]]);
  return 0;
}