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 <map>
#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

typedef long long ll;

ll read() {
	char ch;
	for(ch=getchar();ch<'0'||ch>'9';ch=getchar());
	ll x=ch-'0';
	for(ch=getchar();ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0';
	return x;
}

const int N=1e5+5;

int n,fa[N],ord[N],sz[N];
ll d[N],sum;
vector<int> son[N];
map<ll,int> id;

void dfs(int x,int d) {
	sum+=d;
	for(int y:son[x]) dfs(y,d+1);
}

bool cmp(int x,int y) {return d[x]<d[y];}

int main() {
	n=read();
	fo(i,1,n) d[i]=read(),ord[i]=i,sz[i]=1;
	sort(ord+1,ord+n+1,cmp);
	fo(i,1,n) id[d[ord[i]]]=i;
	fd(i,n,2) {
		int x=ord[i];
		ll s=d[x]-n+2*sz[x];
		if (!id.count(s)) {puts("-1");return 0;}
		int y=id[s];
		if (y>=i) {puts("-1");return 0;}
		fa[x]=ord[y];sz[ord[y]]+=sz[x];
		son[ord[y]].push_back(x);
	}
	dfs(ord[1],0);
	if (sum!=d[ord[1]]) {puts("-1");return 0;}
	fo(i,1,n) if (fa[i]) printf("%d %d\n",i,fa[i]);
	return 0;
}