Problem:
Score : 700 points

Problem Statement

Takahashi and Aoki will play a game with a number line and some segments.
Takahashi is standing on the number line and he is initially at coordinate 0.
Aoki has N segments. The i-th segment is [L_i,R_i], that is, a segment consisting of points with coordinates between L_i and R_i (inclusive).

The game has N steps. The i-th step proceeds as follows:

First, Aoki chooses a segment that is still not chosen yet from the N segments and tells it to Takahashi.

Then, Takahashi walks along the number line to some point within the segment chosen by Aoki this time.

After N steps are performed, Takahashi will return to coordinate 0 and the game ends.

Let K be the total distance traveled by Takahashi throughout the game. Aoki will choose segments so that K will be as large as possible, and Takahashi walks along the line so that K will be as small as possible. What will be the value of K in the end?

Constraints

1 ≤ N ≤ 10^5

-10^5 ≤ L_i < R_i ≤ 10^5

L_i and R_i are integers.

Input

Input is given from Standard Input in the following format:

N
L_1 R_1
:
L_N R_N

Output

Print the total distance traveled by Takahashi throughout the game when Takahashi and Aoki acts as above.
It is guaranteed that K is always an integer when L_i,R_i are integers.

Sample Input 1

3
-5 1
3 7
-4 -2

Sample Output 1

10

One possible sequence of actions of Takahashi and Aoki is as follows:

Aoki chooses the first segment. Takahashi moves from coordinate 0 to -4, covering a distance of 4.

Aoki chooses the third segment. Takahashi stays at coordinate -4.

Aoki chooses the second segment. Takahashi moves from coordinate -4 to 3, covering a distance of 7.

Takahashi moves from coordinate 3 to 0, covering a distance of 3.

The distance covered by Takahashi here is 14 (because Takahashi didn't move optimally).
It turns out that if both players move optimally, the distance covered by Takahashi will be 10.

Sample Input 2

3
1 2
3 4
5 6

Sample Output 2

12

Sample Input 3

5
-2 0
-2 0
7 8
9 10
-2 -1

Sample Output 3

34

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#define ll 					long long
#define db					double
#define ull 				unsigned long long
#define pii 				pair<int,int>
#define fi 					first
#define se 					second
#define vi 					vector<int>
#define psb 				push_back
#define ppb 				pop_back
#define all(x)			 	(x).begin(),(x).end()
#define sz(x)				(int)(x).size()
#define endln 				printf("\n")
#define gc					getchar_unlocked
#define setmin(x)			memset((x), -1, sizeof((x)))
#define setnul(x)			memset((x), 0, sizeof((x)))
#ifndef getchar_unlocked
#define getchar_unlocked 	getchar
#endif
const int inf = 1e9 + 5;
const ll infll = 1e18+5;
const int mod = 1e9 + 7;

template <typename T>
void gi(T &ret) {
	ret = 0; char inp=gc(); int kl=1;
	while (inp<'0' || inp>'9') {if (inp=='-') kl=-1; inp=gc();}
	while ('0'<=inp && inp<='9') ret=(ret<<3)+(ret<<1)+(inp-'0'), inp=gc();
	if (kl<1) ret=-ret;
}

const int MAXN = 1e5+5;
int n;
ll vl[MAXN], vr[MAXN];
bool use[MAXN];
priority_queue <pii> kn;
priority_queue <pii,vector<pii>,greater<pii>> kr;

pii getrm () {
	while (!kn.empty()) {
		pii x = kn.top(); 
		kn.pop();
		if (use[x.se]) continue;
		return x;
	}
	return {-1,-1};
}

pii getlm () {
	while (!kr.empty()) {
		pii x = kr.top();
		kr.pop();
		if (use[x.se]) continue;
		return x;
	}
	return {-1,-1};
}

ll solve () {
	ll ret = 0ll;
	ll now = 0ll;
	setnul(use);
	while (1) {
		pii rv = getrm();
		pii lv = getlm();
		if (rv.se==-1) break;

		// printf ("(%d,%d) - (%d,%d)\n", lv.fi,lv.se, rv.fi, rv.se);

		if (rv.fi<=lv.fi || rv.se==lv.se) {
			int no = rv.se;
			if (now>vr[no]) {
				ret += now-vr[no];
				now = vr[no];
				break;
			}
			no = lv.se;
			if (now<vl[no]) {
				ret += vl[no] - now;
				now = vl[no];
				break;
			}
			break;	
		} else {
			use[lv.se] = use[rv.se] = 1;
			int no = rv.se;
			if (now<vl[no]) {
				ret += vl[no] - now;
				now = vl[no];
			} else if (now>vr[no]) {
				ret += now-vr[no];
				now = vr[no];
			}
			no = lv.se;
			if (now<vl[no]) {
				ret += vl[no] - now;
				now = vl[no];
			} else if (now>vr[no]) {
				ret += now-vr[no];
				now = vr[no];
			}
			// printf ("ret = %lld\n", ret);
		}
	}
	ret += abs(now);
	return ret;
}

int main() {
	gi(n);
	for (int i=1; i<=n; i++) {
		int l, r; gi(l); gi(r);
		kn.push({l,i});
		kr.push({r,i});

		vl[i] = (ll) l;
		vr[i] = (ll) r;
	}
	printf ("%lld\n", solve());
	return 0;
}