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<iostream>
#include<cstring>
#include<cassert>
#include<cmath>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cstdio>
#include<vector>
#include<time.h>
#include<algorithm>
using namespace std;
#define REP(i,x,y) for(ll i=x;i<=y;i++)
#define rep(i,n) REP(i,1,n)
#define rep0(i,n) REP(i,0,n-1)
#define repG(i,x) for(ll i=pos[x];~i;i=e[i].next)
#define ll long long
#define db double
const ll N=1e5+7;
const ll INF=1e9+7;
ll n,cnt=0;
ll sz[N],w[N],X[N],Y[N],ds[N];
struct pir{ll x,d;}p[N];
bool cmp(pir a,pir b){return a.d<b.d;}

int main(){
	scanf("%lld",&n);
	rep(i,n)scanf("%lld",&p[i].d),p[i].x=i;
	sort(p+1,p+n+1,cmp);
	rep(i,n){
		w[i]=p[i].d;
		sz[i]=1;
		ds[i]=0;
	}
	for(ll i=n;i>1;i--){
		ll t=lower_bound(w+1,w+n+1,w[i]-(n-2*sz[i]))-w;
		if(w[t]!=w[i]-(n-2*sz[i])){
			puts("-1");
			return 0;
		}
		sz[t]+=sz[i];
		ds[t]=ds[t]+ds[i]+sz[i];
		X[++cnt]=p[i].x;
		Y[cnt]=p[t].x;
	}
	if(sz[1]!=n||ds[1]!=w[1]){
		puts("-1");
		return 0;
	}
	rep(i,cnt)printf("%lld %lld\n",X[i],Y[i]);
	return 0;
}
