Problem: 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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;

int gae[100001];
vector< pair<long long, int> > p;
vector< pair<int, int> > ans;
map<long long, int> mp;
long long ll[100001];
long long su[100001];

int main(){
    int n;
    scanf("%d", &n);
    for(int i=1;i<=n;i++){
        long long x;
        scanf("%lld", &x);
        p.push_back({x, i});
        gae[i] = 1;
        mp[x] = i;
        su[i] = x;
    }
    sort(p.begin(), p.end());
    for(int i=n-1;i>=1;i--){
        int target = p[i].second;
        long long cal = p[i].first - n + gae[target] + gae[target];
        int nxt = mp[cal];
        //printf("%d %lld\n", nxt, cal);
        if(nxt == 0 || cal >= p[i].first){
            printf("-1\n");
            return 0;
        }
        gae[nxt] += gae[target];
        ll[nxt] += ll[target]; ll[nxt] += gae[target];
        ans.push_back({nxt, target});
    }
    if(ll[p[0].second] == su[p[0].second]){
        for(auto &anss: ans){
            printf("%d %d\n", anss.first, anss.second);
        }
    }else{
        printf("-1\n");
    }
}