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>
#define ll long long
using namespace std;
ll n,sum,a[500001],b[500001],c[500001],sz[500001],id[500001];
vector<ll>zh[150001];
map<ll,ll>mp;
bool cmp(ll x,ll y){
	return a[x]>a[y];
}
void dfs(ll x,ll fa,ll now){
	ll i;
	sum+=now;
	for(i=0;i<zh[x].size();i++)if(zh[x][i]!=fa)dfs(zh[x][i],x,now+1);
}
int main(){
	ll i,x,y,w;
	scanf("%lld",&n);
	for(i=1;i<=n;i++)scanf("%lld",&a[i]),mp[a[i]]=i;
	for(i=1;i<=n;i++)sz[i]=1,id[i]=i;
	sort(id+1,id+n+1,cmp);
	for(i=1;i<n;i++){
		x=id[i];
		w=a[x]-(n-sz[x])+sz[x];
		//prllf("%d %d %d %d\n",x,a[x],sz[x],w);
		if(!mp[w]||w>=a[x])return puts("-1"),0;
		y=mp[w];
		b[i]=x;c[i]=y;
		zh[x].push_back(y);
		zh[y].push_back(x);
		sz[y]+=sz[x];
	}
	dfs(id[n],0,0);
	if(sum!=a[id[n]])return puts("-1"),0;
	for(i=1;i<n;i++)printf("%lld %lld\n",b[i],c[i]);
}
/*
*/