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;
#define int long long
const int N=100005;
map<int ,int > s;
int sd[N],fa[N],rt,cnt[N],ne[N],tot,size[N],f[N],fi[N],zz[N],a[N],n;
int cmp(int x,int y){
	return a[x]>a[y];
}
void dfs(int x,int z){
	sd[x]=z;
	for (int i=fi[x];i;i=ne[i])dfs(zz[i],z+1);
}
void dfs2(int x){
	if (x!=rt)cnt[x]=cnt[fa[x]]+n-2*size[x];
	for (int i=fi[x];i;i=ne[i])dfs2(zz[i]);
}
void jb(int x,int y){
	ne[++tot]=fi[x];
	fi[x]=tot;
	zz[tot]=y;
}
signed main(){
	scanf("%lld",&n);
	for (int i=1;i<=n;i++)scanf("%lld",&a[i]),s[a[i]]=i,f[i]=i;
	sort(f+1,f+n+1,cmp);
	for (int i=1;i<=n;i++){
		size[f[i]]++;
		if (i==n)continue;
		if (!s.count(a[f[i]]-(n-2*size[f[i]]))){
			puts("-1");
			return 0;
		}
		int k=a[f[i]]-(n-2*size[f[i]]);
		size[s[k]]+=size[f[i]];fa[f[i]]=s[k];
		jb(s[k],f[i]);
	}
	rt=0;
	for (int i=1;i<=n;i++)
		if (!fa[i])dfs(i,0),rt=i;
	for (int i=1;i<=n;i++)cnt[rt]+=sd[i];	
	dfs2(rt);
	for (int i=1;i<=n;i++)
		if (a[i]!=cnt[i]){
			puts("-1");
			return 0;
		}	
	for (int i=1;i<=n;i++)
		if (fa[i])printf("%lld %lld\n",i,fa[i]);	
}