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 int long long
using namespace std;
const int N=1e5+10;
int n,cnt,e,f[N],a[N],id[N],w[N],siz[N],res[N],head[N],link[N],to[N];
bool vis[111111];
bool cmp(int x,int y){return a[x]<a[y];}
void add(int u,int v){to[++e]=v,link[e]=head[u],head[u]=e;}
void dfs1(int u,int dep){
	siz[u]=1,res[id[1]]+=dep;
	for(int i=head[u];i;i=link[i])if(to[i]!=f[u])dfs1(to[i],dep+1),siz[u]+=siz[to[i]];
}
void dfs2(int u){for(int i=head[u];i;i=link[i])if(to[i]!=f[u])res[to[i]]=res[u]+n-2*siz[to[i]],dfs2(to[i]);}
signed main(){
	scanf("%lld",&n),w[0]=-LLONG_MAX;
	for(int i=1;i<=n;i++)scanf("%lld",&a[i]),id[i]=i,w[i]=a[i],siz[i]=1;
	sort(id+1,id+n+1,cmp),sort(w+1,w+n+1);
	for(int i=n;i>1;i--){
		int x=2*siz[id[i]]-n+a[id[i]],pos=lower_bound(w+1,w+n+1,x)-w;
		if(w[pos]!=x||n-2*siz[id[i]]<=0){printf("-1");return 0;}
		add(id[pos],id[i]),siz[id[pos]]+=siz[id[i]],f[id[i]]=id[pos];
	}
	dfs1(id[1],0),dfs2(id[1]);
	for(int i=1;i<=n;i++)if(a[i]!=res[i]){printf("-1");return 0;}
	for(int i=2;i<=n;i++)printf("%lld %lld\n",f[id[i]],id[i]);
}