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;

const int MAX_N = 2e5 + 10;

int N, rnk[MAX_N], siz[MAX_N], pa[MAX_N], depth[MAX_N];
ll D[MAX_N];
map<ll, int> mp;

int main() {
    scanf("%d", &N);
    for (int i = 1; i <= N; i++) scanf("%lld", &D[i]), mp[D[i]] = i;
    for (int i = 1; i <= N; i++) rnk[i] = i, siz[i] = 1;
    sort(rnk + 1, rnk + N + 1, [](int a, int b) { return D[a] > D[b]; });
    for (int i = 1; i < N; i++) {
        int x = rnk[i];
        ll pd = D[x] + 2 * siz[x] - N;
        if (siz[x] * 2 < N && mp.count(pd)) {
            pa[x] = mp[pd];
            siz[pa[x]] += siz[x];
        } else {
            printf("-1\n");
            return 0;
        }
    }
    ll sum = 0;
    for (int i = N - 1; i; i--) {
        int x = rnk[i];
        depth[x] = depth[pa[x]] + 1;
        sum += depth[x];
    }
    if (sum != D[rnk[N]]) {
        printf("-1\n");
        return 0;
    }
    for (int i = 1; i < N; i++) {
        int x = rnk[i];
        printf("%d %d\n", x, pa[x]);
    }
}
