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;

const int maxn = 1e5 + 7;

typedef long long LL;

struct Edge{
	int to,last;
}E[maxn];
LL d[maxn];
unordered_map<LL,int>M_T;

int n;
int cnt;
int siz[maxn];
int head[maxn];
LL sum;
void Adde(int x,int y){
	E[++cnt] = (Edge){y,head[x]};
	head[x] = cnt;
}
void Dfs(int x,LL cur){
	sum += cur;
	for(int i = head[x],y; y = E[i].to,i; i = E[i].last){
		Dfs(y,cur + 1);
	}
}
void Print(int x){
	for(int i = head[x],y; y = E[i].to,i; i = E[i].last){
		printf("%d %d\n",x,y);
		Print(y);
	}
}
int main(){
	scanf("%d",&n);
	for(int i = 1; i <= n; ++i)scanf("%lld",&d[i]),M_T[d[i]] = i,siz[i] = 1;
	sort(d + 1,d + n + 1);
	for(int i = n; i > 1; --i){
		int x = M_T[d[i]];
		int tmp = n - 2 * siz[x];
		if(tmp <= 0)return !puts("-1");
		LL cur = d[i] - tmp;
		if(!M_T.count(cur))return !puts("-1");
		int prt = M_T[cur];
		Adde(prt,x);
		siz[prt] += siz[x];
	}
	Dfs(M_T[d[1]],0);
	if(sum != d[1])return !puts("-1");
	Print(M_T[d[1]]);
	return 0;
}