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 maxn 100003
using namespace std;
struct edge{int to,next;}e[maxn<<1];
int head[maxn],cnte,fa[maxn];
void add(int u,int v){e[++cnte].to=v,e[cnte].next=head[u],head[u]=cnte;}
int n,sz[maxn],s[maxn];
long long d[maxn],sum;
map<long long,int> b;
bool cmp(int x,int y){return d[x]<d[y];}
void dfs(int u,int depth){
	sum+=depth;
	for(int i=head[u];i;i=e[i].next){
		int v=e[i].to;
		if(v==fa[u])continue;
		dfs(v,depth+1);
	}
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%lld",d+i),s[i]=i,b[d[i]]=i,sz[i]=1;
	sort(s+1,s+n+1,cmp);
	for(int i=n;i>=2;i--){
		int v=s[i];
		long long d_u=d[v]-(n-sz[v])+sz[v];
		if(d_u>=d[v]||!b.count(d_u)){puts("-1");return 0;}
		int u=b[d_u];
		add(u,v),add(v,u),fa[v]=u;
		sz[u]+=sz[v];
	}
	if(cnte!=(n-1)*2){puts("-1");return 0;}
	dfs(s[1],0);
	if(sum!=d[s[1]]){puts("-1");return 0;}
	for(int u=1;u<=n;u++){
		for(int i=head[u];i;i=e[i].next){
			int v=e[i].to;
			if(v==fa[u])continue;
			printf("%d %d\n",u,v);
		}
	}
	return 0;
}
//sykakioi