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 pair<int, int> pii;
typedef long long li;

const int maxn = 1e5;

int n, fa[maxn];
li D[maxn];
int size[maxn];

pair<li, int> ord[maxn];
map<li, int> id;
li dp[maxn];

int main(void) {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%lld", D + i);
  for (int i = 0; i < n; ++i) ord[i] = {D[i], i};
  for (int i = 0; i < n; ++i) id[D[i]] = i;
  sort(ord, ord + n);
  reverse(ord, ord + n);
  for (int i = 0; i < n; ++i) size[i] = 1;
  for (int i = 0; i < n; ++i) dp[i] = 0;

  for (int i = 0; i < n - 1; ++i) {
    int u = ord[i].second;
    li del = n - size[u] * 2;
    if (del <= 0 || !id.count(D[u] - del)) {
      puts("-1");
      return 0;
    }
    fa[u] = id[D[u] - del];
    size[fa[u]] += size[u];
    dp[fa[u]] += dp[u] + size[u];
  }

  if (dp[ord[n - 1].second] != D[ord[n - 1].second]) {
    puts("-1");
    return 0;
  }

  for (int i = 0; i < n - 1; ++i) {
    int u = ord[i].second;
    int v = fa[u];
    printf("%d %d\n", u + 1, v + 1);
  }
  return 0;
}
