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;
typedef pair<int, int> ii;

const int N = 1e6 + 5;

vector<int> adj[N];
int n, sub[N];
ll d[N], D;
map<ll, int> pos;

void no_answer() {
  puts("-1");
  exit(0);
}

void calc(int u, int p, int d = 0) {
  D += d; 
  for (int v : adj[u]) calc(v, u, d + 1);
}

void check(int u, int p, ll D) {
  if (D != d[u]) no_answer();
  for (int v : adj[u]) check(v, u, D + n - 2 * sub[v]);
}

int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", &d[i]);
    pos[d[i]] = i;
  }

  vector<int> ord(n);
  iota(ord.begin(), ord.end(), 1);
  sort(ord.begin(), ord.end(), [](int i, int j) { return d[i] > d[j]; });

  for (int i = 0; i < n - 1; ++i) {
    int u = ord[i];
    ++sub[u];
    ll dp = d[u] - n + 2 * sub[u];
    if (!pos.count(dp)) no_answer();
    int p = pos[dp];
    sub[p] += sub[u];
    adj[p].push_back(u);
  }

  calc(ord[n - 1], 0);
  check(ord[n - 1], 0, D);

  for(int i = 1; i <= n; ++i) {
    for(int j : adj[i]) {
      printf("%d %d\n", i, j); 
    }
  }

  return 0;
}
