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>
using namespace std;
const int N=1e5+100;
typedef long long ll;
int n,u[N],v[N],sz[N],cnt,RT;
int head[N],to[N],nxt[N],tot;
ll wd[N];

struct P{ll v;int id;}t[N];
inline bool cmp(const P&A,const P&B)
{return B.v<A.v;}

map<ll,int>d;

inline void lk(int u,int v)
{to[++tot]=v;nxt[tot]=head[u];head[u]=tot;}

void dfs(int x,int dep)
{
	wd[RT]+=dep;
	for(int i=head[x];i;i=nxt[i]) dfs(to[i],dep+1);
}

void dfss(int x){
  for(int i=head[x];i;i=nxt[i]) 
   {wd[to[i]]=wd[x]+n-(sz[to[i]]<<1);dfss(to[i]);}
}

int main(){
	int i,j;ll dr;
	scanf("%d",&n);
	for(i=1;i<=n;++i){
		scanf("%lld",&t[i].v);
		d[t[i].v]=i;t[i].id=i;
		sz[i]=1;
	}
	sort(t+1,t+n+1,cmp);
	for(i=1;i<n;++i){
		dr=t[i].v-n+(sz[t[i].id]<<1);
		if(dr<0 || (sz[t[i].id]<<1)==n || (!d[dr])) {puts("-1");return 0;}
		j=d[dr];u[++cnt]=j;v[cnt]=t[i].id;
		sz[j]+=sz[t[i].id];
	}
	for(i=1;i<n;++i) lk(u[i],v[i]);
	RT=t[n].id;
	dfs(RT,0);dfss(RT);
	for(i=1;i<=n;++i){
		if(wd[t[i].id]!=t[i].v) {puts("-1");return 0;}
	}
	for(i=1;i<n;++i) printf("%d %d\n",u[i],v[i]);
	return 0;
}