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;
struct node
{
    ll d;int id;
}c[100005];
bool cmp(node x,node y){return x.d>y.d;}
int fa[100005],siz[100005],deep[100005];
ll d[100005];
vector<int>to[100005];
void dfs(int x){for(auto tar:to[x])deep[tar]=deep[x]+1,dfs(tar);}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%lld",&c[i].d),c[i].id=i,siz[i]=1;
    sort(c+1,c+n+1,cmp);
    for(int i=1;i<=n;i++)d[i]=c[n+1-i].d;
    for(int i=1;i<n;i++)
    {
        int t=lower_bound(d+1,d+n+1,c[i].d+2*siz[c[i].id]-n)-d;
        t=n+1-t;
        if(c[t].d!=c[i].d+2*siz[c[i].id]-n){puts("-1");return 0;}
        fa[c[i].id]=c[t].id,siz[c[t].id]+=siz[c[i].id];
    }
    for(int i=1;i<=n;i++)if(fa[i])to[fa[i]].push_back(i);
    dfs(c[n].id);
    ll s=0;
    for(int i=1;i<=n;i++)s+=deep[i];
    if(s!=c[n].d){puts("-1");return 0;}
    for(int i=1;i<=n;i++)if(fa[i])printf("%d %d\n",i,fa[i]);
    return 0;
}