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 std::cerr;
using std::endl;

typedef long long ll;

inline ll rd(){
  ll x; scanf("%lld",&x);
  return x;
}

const int N=1e5+233;

int n;
ll D[N],S[N];
std::vector<std::pair<int,int>>ans;
std::map<ll,int>map;
std::vector<int> T[N];

ll dfs(int x,int fa,int dis){
  ll ret=dis;
  for(int y:T[x]) if(y!=fa)
    ret+=dfs(y,x,dis+1);
  return ret;
}

int main(){
  //freopen("data","r",stdin);
  n=rd();
  for(int i=1;i<=n;++i){
    D[i]=rd();
    S[i]=1;
    map[-D[i]]=i;
  }
  while(!map.empty()){
    if(map.size()==1)
      break;
    int x=map.begin()->second;
    map.erase(map.begin());
    std::map<ll,int>::iterator it;
    it=map.find(-(D[x]-n+2*S[x]));
    if(it==map.end()){
      puts("-1");
      return 0;
    }
    S[it->second]+=S[x];
    ans.push_back({x,it->second});
    T[x].push_back(it->second);
    T[it->second].push_back(x);
  }
  int root=map.begin()->second;
  if(dfs(root,0,0)!=D[root]){
    puts("-1");
    return 0;
  }
  for(int i=0;i<n-1;++i)
    printf("%d %d\n",ans[i].first,ans[i].second);
  return 0;
}
