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 <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>

#define int long long

using namespace std;

int n, S[100010], W[100010], D[100010], Sz[100010], F[100010], ans = 0;

int find(int x) {
    int t = lower_bound(D + 1, D + n + 1, x) - D;
    if(D[t] != x) {
        puts("-1");
        exit(0);
    }
    return t;
}

signed main() {
    scanf("%lld", &n);
    for(int i = 1; i <= n; i++) scanf("%lld", &D[i]);
    memcpy(S, D, sizeof D);
    sort(D + 1, D + n + 1);
    for(int i = 1; i <= n; i++) W[find(S[i])] = i;
    for(int i = 1; i <= n; i++) Sz[i] = 1;
    for(int i = n; i > 1; i--) {
        F[i] = find(D[i] - n + Sz[i] * 2);
        Sz[F[i]] += Sz[i];
        ans += Sz[i];
    }
    if(ans != D[1]) {
        puts("-1");
        return 0;
    }
    for(int i = 2; i <= n; i++) printf("%lld %lld\n", W[i], W[F[i]]);
    return 0;
}