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 X first
#define Y second
using namespace std;

typedef long long ll;
typedef pair<ll,ll> pll;
const ll N = 100005;

ll n, d[N], cnt[N], rd[N];
map<ll,ll> dti, itd;
vector<ll> adj[N];
vector<pll> ans;

void dfs1 (ll I) {
	for(auto &T : adj[I]) {
		dfs1(T);
		rd[I] += rd[T] + cnt[T];
	}
}

void dfs2 (ll I) {
	for(auto &T : adj[I]) {
		rd[T] = rd[I] - cnt[T] + (n-cnt[T]);
		dfs2(T);
	}
}

int main()
{
	scanf("%lld",&n);
	for(ll i=1;i<=n;i++) {
		scanf("%lld",&d[i]);
		dti[d[i]] = i;
		itd[i] = d[i];
	}
	sort(d+1, d+1+n);
	for(ll i=n;i>=2;i--) {
		ll I = dti[d[i]];
		ll C = ++cnt[I];
		ll J = dti[d[i] - (n-C) + C];
		if(J == 0) {
			puts("-1");
			return 0;
		}
		ans.push_back({I, J});
		cnt[J] += cnt[I];
		adj[J].push_back(I);
	}
	ll R = dti[d[1]];
	dfs1(R);
	dfs2(R);
	for(ll i=1;i<=n;i++) {
		if(itd[i] != rd[i]) {
			puts("-1");
			return 0;
		}
	}
	for(auto &T : ans) {
		printf("%lld %lld\n", T.X, T.Y);
	}
}
