#define _USE_MATH_DEFINES
#include <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <string>
#include <vector>
#include <utility>
#include <complex>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <deque>
#include <tuple>
#include <bitset>
#include <limits>
#include <algorithm>
#include <array>
#include <random>
#include <complex>
#include <regex>
using namespace std;
typedef long double ld;
typedef long long ll;
typedef vector<int> vint;
typedef vector<ll> vll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
typedef complex<ld> compd;
#define quickIO()	{cin.tie(0);	cout.sync_with_stdio(false);}
#define reach(i,a)	for(auto i:a)
#define rep(i,n)	for(int i=0;i<((int)n);i++)
#define REP(i,n)	for(int i=0;i<=((int)n);i++)
#define srep(i,a,n)	for(int i=a;i<((int)n);i++)
#define SREP(i,a,n)	for(int i=a;i<=((int)n);i++)
#define rrep(i,n)	for(int i=n-1;i>=0;i--)
#define RREP(i,n)	for(int i=n;i>=0;i--)
#define all(a)	(a).begin(),(a).end()
#define mp(a,b)	make_pair(a,b)
#define mt	make_tuple
#define pb	push_back
template<typename T> istream& operator >> (istream& is, vector<T>& vec) {
	for (T& x : vec)	is >> x;
	return is;
}
template<typename T> ostream& operator << (ostream& os, vector<T>& vec) {
	os << "[";
	rep(i, vec.size())	os << (i ? ", " : "") << vec[i];
	os << "]";
	return os;
}
template<typename T> istream& operator >> (istream& is, pair<T, T>& p) {
	is >> p.first >> p.second;
	return is;
}
template<typename T> ostream& operator << (ostream& os, pair<T, T>& p) {
	os << "(" << p.first << ", " << p.second << ")";
	return os;
}
int bitcnt(ll x) {
	x = ((x & 0xAAAAAAAAAAAAAAAA) >> 1) + (x & 0x5555555555555555);
	x = ((x & 0xCCCCCCCCCCCCCCCC) >> 2) + (x & 0x3333333333333333);
	x = ((x & 0xF0F0F0F0F0F0F0F0) >> 4) + (x & 0x0F0F0F0F0F0F0F0F);
	x = ((x & 0xFF00FF00FF00FF00) >> 8) + (x & 0x00FF00FF00FF00FF);
	x = ((x & 0xFFFF0000FFFF0000) >> 16) + (x & 0x0000FFFF0000FFFF);
	x = ((x & 0xFFFFFFFF00000000) >> 32) + (x & 0x00000000FFFFFFFF);
	return x;
}
int bitcnt(int x) {
	x = ((x & 0xAAAAAAAA) >> 1) + (x & 0x55555555);
	x = ((x & 0xCCCCCCCC) >> 2) + (x & 0x33333333);
	x = ((x & 0xF0F0F0F0) >> 4) + (x & 0x0F0F0F0F);
	x = ((x & 0xFF00FF00) >> 8) + (x & 0x00FF00FF);
	x = ((x & 0xFFFF0000) >> 16) + (x & 0x0000FFFF);
	return x;
}
ll sqrtll(ll x) {
	ll left = 0, right = x;
	rep(i, 100) {
		ll mid = (left + right) >> 1;
		if (mid*mid <= x)	left = mid;
		else	right = mid;
	}
	return left;
}
ll gcd(ll a, ll b) {
	return a % b == 0 ? b : gcd(b, a%b);
}
#define debug(x)	printf("Case #%d: ", x)
#define DEBUG 0
const ll inf = 1e18;
const ld infl = 1e100;
const ll mod = 1e9+7;
const ld eps = 1e-9;
const int dx[] = { 1,0,-1,0,0 };
const int dy[] = { 0,1,0,-1,0 };

int group[100010];

int getGroup(int i) {
	return group[i] = (i == group[i] ? group[i] : getGroup(group[i]));
}
void merge(int i, int j) {
	i = getGroup(i);	j = getGroup(j);
	group[i] = group[j];
}

// 最初の2頂点はxとx-n+2があったらマージ。headのみを残す。
// 以降x(a)とy(b)についてx+n-2*b=yやy+n-2*a=xで判断
// x(a)は固定できるけど、y(b)が見つけられない
// xが親になる場合はx+n=y+2*b->子になる場合はx+2*a-n=y
// 「一番大きいdiは子になる」より再帰的に(UF?)出来そう

// さらに、全体を+sしても上の状態では成り立ってしまうので、実際の距離と一致しているか検証する必要がある。

int main() {
	ll n;	cin >> n;
	vll d(n);	cin >> d;
	vll num(n, 1);
	map<ll, ll> mp;
	rep(i, n) {
		mp[d[i]] = i;
	}
	vector<pll> d2(n);
	rep(i, n) {
		d2[i] = mp(d[i], i);
		group[i] = i;
	}
	sort(all(d2));
	vector<pll> ret;
	for (ll i = n - 1; i > 0; i--) {
		ll id = d2[i].second;
		if (getGroup(id) != id)	continue;
		ll a = num[id];
		if (!mp.count(d[id] + 2 * a - n) || id == mp[d[id] + 2 * a - n]) {
			cout << -1 << endl;
			return 0;
		}
		ll to = mp[d[id] + 2 * a - n];
		mp.erase(d[id]);
		merge(id, to);
		num[to] += num[id];
		ret.push_back(mp(id + 1, to + 1));
	}
	// 検証パート
	num = vll(n, 1);
	vll dist = vll(n, 0);
	rep(i, n - 1) {
		dist[ret[i].second - 1] += dist[ret[i].first - 1] + num[ret[i].first - 1];
		num[ret[i].second - 1] += num[ret[i].first - 1];
	}
	ll root = ret[n - 2].second - 1;
	if (dist[root] != d[root]) {
		cout << -1 << endl;
		return 0;
	}
	rep(i, n - 1) {
		cout << ret[i].first << " " << ret[i].second << endl;
	}
	return 0;
}

Measured memory_bytes: 12152