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
//ios::sync_with_stdio(false);
#include<bits/stdc++.h>
#define LL long long
#define F(x,y,z) for(int x=y;x<=z;++x)
#define D(x,y,z) for(int x=y;x>=z;--x)
#define ckmax(a,b) (a)=max(a,b)
#define ckmin(a,b) (a)=min(a,b)
#define pb push_back
using namespace std;
const int N=100010;
LL R(){
    LL ans=0,f=1;char c=getchar();
    for(;c<'0'||c>'9';c=getchar()) if (c=='-') f=-1;
    for(;c>='0'&&c<='9';c=getchar()) ans=ans*10+c-'0';
    return ans*f;
}
int n,fa[N],siz[N];
vector<int> v[N];
LL d[N],s,g[N];
map<LL,int> ys,id;
void Err(){
	printf("-1\n");
	exit(0);
}
void dfs(int u,int dep){
	s+=dep;
	for(auto to:v[u])dfs(to,dep+1);
}
int main(){
	//freopen("1.in","r",stdin);freopen("1.out","w",stdout);
	n=R();
	F(i,1,n)d[i]=R(),siz[i]=1,id[d[i]]=i;
	sort(d+1,d+1+n);
	F(i,1,n)ys[d[i]]=i;
	D(i,n,2){
		fa[i]=ys[d[i]+siz[i]*2-n];
		if(fa[i]==0)Err();
		v[fa[i]].pb(i);
		siz[fa[i]]+=siz[i];
	}
	dfs(1,0);
	if(s!=d[1])Err();
	F(i,2,n)printf("%d %d\n",id[d[i]],id[d[fa[i]]]);
    return 0;
}