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>
#define fo(i, x, y) for(int i = x, B = y; i <= B; i ++)
#define ff(i, x, y) for(int i = x, B = y; i <  B; i ++)
#define fd(i, x, y) for(int i = x, B = y; i >= B; i --)
#define ll long long
#define pp printf
#define hh pp("\n")
using namespace std;

const int N = 1e5 + 5;

int n;
ll d[N];
int a[N];

map<ll, int> bz;

int cmpa(int x, int y) {
	return d[x] > d[y];
}

int ans[N][2], ans0;
vector<int> e[N];

int siz[N];

void Noans() {
	pp("-1\n");
	exit(0);
}

ll sum;

void dg(int x, int y) {
	sum += y;
	ff(j, 0, e[x].size()) {
		dg(e[x][j], y + 1);
	}
}

int main() {
	scanf("%d", &n);
	fo(i, 1, n) scanf("%lld", &d[i]), bz[d[i]] = i, a[i] = i;
	sort(a + 1, a + n + 1, cmpa);
	fo(i, 1, n) siz[i] = 1;
	fo(i, 1, n - 1) {
		int x = a[i];
		if(siz[x] * 2 > n)
			Noans();
		ll y = d[x] - n + 2 * siz[x];
		if(!bz.count(y)) Noans();
		y = bz[y];
		siz[y] += siz[x];
		ans[++ ans0][0] = x, ans[ans0][1] = y;
		e[y].push_back(x);
	}
	dg(a[n], 0);
	if(sum != d[a[n]]) Noans();
	fo(i, 1, n - 1) pp("%d %d\n", ans[i][0], ans[i][1]);
}