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<cstdio>
#include<algorithm>
using namespace std;
const int N=100010;
int n,l,r,mid,j,p[N],siz[N],fa[N];
long long d[N],dis[N];
bool cmp(int x,int y) {return d[x]<d[y];}
long long read() {
	long long tmp=0, fh=1; char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') fh=-1; c=getchar();}
	while (c>='0'&&c<='9') tmp=tmp*10+c-48, c=getchar();
	return tmp*fh;
}
int main() {
	n=read();
	for (int i=1;i<=n;i++) d[i]=read(), p[i]=i, siz[i]=1;
	sort(p+1,p+n+1,cmp);
	for (int i=n,j=i-1;i>1;i--) {
		l=1; r=i;
		while (l+1<r) {
			mid=(l+r)>>1;
			if (d[p[mid]]<=d[p[i]]+siz[p[i]]*2-n) l=mid; else r=mid;
		}
		j=l;
		if (d[p[j]]==d[p[i]]+siz[p[i]]*2-n) siz[p[j]]+=siz[p[i]], dis[p[j]]+=dis[p[i]]+siz[p[i]], fa[p[i]]=p[j];
		else {
			printf("-1\n"); return 0;
		}
	}
	if (dis[p[1]]!=d[p[1]]) printf("-1\n"); else
		for (int i=2;i<=n;i++) printf("%d %d\n",fa[p[i]],p[i]);
	return 0;
} 