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
#define N 500005
using namespace std;
int n,cnt,sum,a[N],sz[N],tmp[N],ansx[N],ansy[N];
map<int,int>b;
vector<int>G[N];
inline bool cmp(int x,int y){return a[x]>a[y];}
void dfs(int u,int fa,int dep){
	sum+=dep;
	for (int i=0;i<(int)G[u].size();i++){
		int v=G[u][i];
		if (v==fa) continue;
		dfs(v,u,dep+1);
	}
}
signed main(){
	scanf("%lld",&n);
	for (int i=1;i<=n;i++) scanf("%lld",&a[i]),b[a[i]]=i,tmp[i]=i,sz[i]=1;
	sort(tmp+1,tmp+n+1,cmp);
	for (int i=1;i<n;i++){
		int x=tmp[i];
		int dis=a[x]-(n-sz[x])+sz[x];
		if (!b[dis]) return puts("-1"),0;
		int y=b[dis];
		sz[y]+=sz[x];ansx[++cnt]=x;ansy[cnt]=y;
		if (x==y) return puts("-1"),0;
		G[y].push_back(x);
	}
	int rt=tmp[n];sum=0;
	dfs(rt,-1,0);
	if (sum!=a[rt]) return puts("-1"),0;
	for (int i=1;i<=cnt;i++) printf("%lld %lld\n",ansx[i],ansy[i]);
	return 0;
}