CodeForces–830A--二分，贪心

Office Keys

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are n people and k keys on a straight line. Every person wants to get to the office which is located on the line as well. To do that, he needs to reach some point with a key, take the key and then go to the office. Once a key is taken by somebody, it couldn't be taken by anybody else.

You are to determine the minimum time needed for all n people to get to the office with keys. Assume that people move a unit distance per 1 second. If two people reach a key at the same time, only one of them can take the key. A person can pass through a point with a key without taking it.

Input

The first line contains three integers n, k and p (1 ≤ n ≤ 1 000, n ≤ k ≤ 2 000, 1 ≤ p ≤ 10⁹) — the number of people, the number of keys and the office location.

The second line contains n distinct integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹) — positions in which people are located initially. The positions are given in arbitrary order.

The third line contains k distinct integers b₁, b₂, ..., b_k (1 ≤ b_j ≤ 10⁹) — positions of the keys. The positions are given in arbitrary order.

Note that there can't be more than one person or more than one key in the same point. A person and a key can be located in the same point.

Output

Print the minimum time (in seconds) needed for all n to reach the office with keys.

Examples

Input

2 4 50
20 100
60 10 40 80

Output

50

Input

1 2 10
11
15 7

Output

7

Note

In the first example the person located at point 20 should take the key located at point 40 and go with it to the office located at point 50. He spends 30 seconds. The person located at point 100 can take the key located at point 80 and go to the office with it. He spends 50 seconds. Thus, after 50 seconds everybody is in office with keys.

思路：

最小化最大值问题

二分，二分最终结果，然后check即可。

check时用到了一个贪心的性质：对某个人来说，若去取某个位于他左边的key满足限制，则取位于这个key右边，p左边的key也满足

代码：

/*
* @Author: zhans
* @Date:   2017-08-18 09:44:32
* @Last Modified by:   zhans
* @Last Modified time: 2017-08-18 14:17:50
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 2000 + 30;
int per[MAXN], key[MAXN];
int vis[MAXN], vis2[MAXN];
int cnt;
int n, k, p;
bool ok(long long mid)
{
	//memset(vis,0,sizeof(vis));
	ll output = 0;
	ll now = 1;
	for (ll i = 1; i <= n; i++)
	{
		while (now <= k && abs(key[now] - per[i]) + abs(key[now] - p) > mid)
		{
			now++;
		}
		if (now <= k)
		{
			now++;
			output++;
		}
		else break;
	}
	if (output >= n)return 1;
	else return 0;
}
int main()
{
	// freopen("data.in", "r", stdin);
	scanf("%d%d%d", &n, &k, &p);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &per[i]);
	}
	for (int i = 1; i <= k; i++) {
		scanf("%d", &key[i]);
	}
	//for(int i=0;i<k;i++) mp[i]=i;
	sort(per+1, per + n+1);
	sort(key+1, key + k+1);
	long long l = 0, r = 2e9;
	while (r - l > 1) {
		long long mid = (l + r) / 2;
		if (ok(mid)) {
			r = mid;
		}
		else {
			l = mid;
		}
	}
	if(ok(l)) r=l;
	printf("%I64d
", r);
}