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>
#define FR first
#define SE second

using namespace std;

typedef long long ll;
typedef pair<ll,int> pr;

map <ll,int> mp;
pr num[100005];

vector <int> e[100005];
int siz[100005],dep[100005];

void dfs(int x) {
  for(int i=0;i<e[x].size();i++) {
  	int u=e[x][i];
  	dep[u]=dep[x]+1;
  	dfs(u);
  }
}

int main() {
  int n;
  scanf("%d",&n);
  for(int i=1;i<=n;i++) {
  	ll x;
  	scanf("%lld",&x);
  	num[i]=pr(x,i);
  	mp[x]=i;
  }
  sort(num+1,num+n+1);
  for(int i=1;i<=n;i++) siz[i]=1;
  for(int i=n;i>1;i--) {
  	int x=num[i].SE;
  	if (siz[x]*2>=n) {
  		puts("-1");
  		return 0;
	  }
	ll v=num[i].FR-(n-2*siz[x]);
	if (!mp.count(v)) {
		puts("-1");
		return 0;
	}
	int p=mp[v];
	e[p].push_back(x);
	siz[p]+=siz[x];
  }
  if (siz[num[1].SE]!=n) {
  	puts("-1");
  	return 0;
  }
  dfs(num[1].SE);
  ll s=0;
  for(int i=1;i<=n;i++) s+=dep[i];
  if (s!=num[1].FR) {
  	puts("-1");
  	return 0;
  }
  for(int i=1;i<=n;i++)
    for(int j=0;j<e[i].size();j++) printf("%d %d\n",i,e[i][j]);
  return 0;
}