Problem: 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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=100005,p=1000000007;
int read(){
	int f=1,g=0;
	char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-1;
	for (;isdigit(ch);ch=getchar()) g=g*10+ch-'0';
	return f*g;
}
int n,fl,siz[N],b[N],fa[N];
vector<int> e[N];
ll a[N],f[N],ans[N];
unordered_map<ll,int> h;
bool cmp(int x,int y){return a[x]<a[y];}
void work(int x){
	f[x]=siz[x]-1;
	for (auto y : e[x]){work(y);f[x]+=f[y];}
}
void calc(int x,ll s){
	ans[x]=f[x]+s+n-siz[x];
	for (auto y : e[x])
	calc(y,ans[x]-f[y]-siz[y]);
}
int main(){
	n=read();
	for (int i=1;i<=n;i++) {scanf("%lld",a+i);h[a[i]]=i;b[i]=i;siz[i]=1;}
	sort(b+1,b+n+1,cmp);
	for (int i=n;i>1;i--){
		int x=b[i];
		fa[x]=h[a[x]+siz[x]-n+siz[x]];
		if (!fa[x]){puts("-1");return 0;}
		siz[fa[x]]+=siz[x];
		e[fa[x]].push_back(x);
	}
	work(b[1]);
	calc(b[1],0);
	for (int i=1;i<=n;i++)
	if (ans[i]!=a[i]){puts("-1");return 0;}
	for (int i=2;i<=n;i++) printf("%d %d\n",fa[b[i]],b[i]);
	return 0;
}