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 <algorithm>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <map>

typedef long long LL;

LL read() {
  LL ans = 0; int c, f = 1;
  while (!isdigit(c = getchar()))
    if (c == '-') f *= -1;
  do ans = ans * 10 + c - '0';
  while (isdigit(c = getchar()));
  return ans * f;
}

const int N = 100050;
std::map<LL, int> M;

int n, siz[N], fa[N];
LL p[N];

int main() {
  n = read();
  for (int i = 1; i <= n; ++i)
    siz[M[read()] = i] = 1;
  for (std::map<LL, int>::iterator it = --M.end(); it != M.begin(); --it) {
    LL D = it->first;
    int i = it->second;
    LL t = D - n + 2 * siz[i];
    if (t >= D || !M.count(t)) { puts("-1"); return 0; }
    int j = M[t];
    siz[fa[i] = j] += siz[i];
    p[j] += siz[i] + p[i];
  }
  if (p[M.begin()->second] != M.begin()->first) {
    puts("-1"); return 0;
  }
  for (int i = 1; i <= n; ++i) if (fa[i])
    printf("%d %d\n", fa[i], i);
}
