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 <algorithm>
#include <cstdio>
#include <vector>

#define pb push_back
#define mp make_pair
#define fi first
#define se second

typedef long long ll;

const int N=1e5+5;

int n,sz[N]; ll d_rt;

std::vector<int> g[N];

std::pair<ll,int> a[N];

void dfs(int u,int dep){
	d_rt+=dep; for(auto v: g[u]) dfs(v,dep+1);
}

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i){ scanf("%lld",&a[i].fi); a[i].se=i; sz[i]=1; }
	std::sort(a+1,a+1+n);
	for(int i=n;i>=2;--i){
		int u=a[i].se; ll d_fa=a[i].fi+2*sz[u]-n;
		int pos=std::lower_bound(a+1,a+1+n,std::mp(d_fa,0))-a;
		if(a[pos].fi!=d_fa){ puts("-1"); return 0; }
		sz[a[pos].se]+=sz[u]; g[a[pos].se].pb(u);
	}
	dfs(a[1].se,0); if(d_rt!=a[1].fi){ puts("-1"); return 0; }
	for(int u=1;u<=n;++u){
		std::sort(g[u].begin(),g[u].end());
		for(auto v: g[u]) printf("%d %d\n",u,v);
	}
	return 0;
}