E. Read Time

E. Read Time

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data in parallel.

When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the i-th reading head is above the track number hi. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered h1, h2, ..., hn have been read at the beginning of the operation.

Mike needs to read the data on m distinct tracks with numbers p1, p2, ..., pm. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read.

Input

The first line of the input contains two space-separated integers n, m (1 ≤ n, m ≤ 105) — the number of disk heads and the number of tracks to read, accordingly. The second line contains n distinct integers hi in ascending order (1 ≤ hi ≤ 1010, hi < hi + 1) — the initial positions of the heads. The third line contains m distinct integers pi in ascending order (1 ≤ pi ≤ 1010, pi < pi + 1) - the numbers of tracks to read.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is recommended to use the cin, cout streams or the %I64d specifier.

Output

Print a single number — the minimum time required, in seconds, to read all the needed tracks.

Examples

input

3 42 5 61 3 6 8

output

2

input

3 31 2 31 2 3

output

0

input

1 2165142 200

output

81

Note

The first test coincides with the figure. In this case the given tracks can be read in 2 seconds in the following way:

1. during the first second move the 1-st head to the left and let it stay there;
2. move the second head to the left twice;
3. move the third head to the right twice (note that the 6-th track has already been read at the beginning).

One cannot read the tracks in 1 second as the 3-rd head is at distance 2 from the 8-th track.

题意：第一行磁头个数和待读取文件的个数。

接下来一行是n个磁头的位置。

接下来一行是m个待读取文件的位置。

然后你可以磁头可以左右移动，问你读取完这些文件最少需要的时间是多少。

不难想到利用二分可以固定当前时间，固定当前时间之后你需要做的就是让这n个磁头把尽可能多的文件给读取掉。

这里采用的策略就是当前文件位置l，然后不断扩大区间，一直到这个磁头不能完全读完这n个文件为止。

那么，判断条件是什么呢，学过操作系统的都知道，最优的磁头走法就是先到达离当前磁头最近的一个端点处，然后扫一次区间，这样用的时间是最少的。

之后的看代码就懂了。

#include <cstdio>#include <queue>#include <iostream>#include <string>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const long long inf = 1e18;const int MAXN = 1e5+7;int n,m;long long h[MAXN],p[MAXN];bool check(long long mid){    int j = 0;    for(int i = 0 ;i < n ; ++i)    {        long long t = mid;        long long pos = h[i];        int l = j;        while(p[j] - p[l] + abs(p[j] - pos) <= t || p[j] - p[l] + abs(p[l] - pos) <= t )        {            j++;            if(j == m)return 1;        }    }    return 0;}int main(){    scanf("%d%d",&n,&m);    for(int i = 0 ; i < n ; ++i)scanf("%I64d",&h[i]);    for(int i = 0 ; i < m ; ++i)scanf("%I64d",&p[i]);    long long low = 0 ,high = 2e10,ans;    while(low <= high)    {        long long mid = (low+high)>>1ll;        if(check(mid))        {            ans = mid;            high = mid - 1;        }        else low = mid + 1;    }    cout << ans <<endl;    return 0;}