/*

怎么判断能否在时间k内完成扫描

贪心：每次取出最靠左边的磁头去扫描最左边的，然后再往右扫描即可

如果当前点无法扫到最左侧点，那么后继点一样无法扫到

*/

#include<bits/stdc++.h>

#define maxn 100005

#define ll long long

using namespace std;

int n,m;

ll h[maxn],p[maxn];

int judge(ll x){

    ll time1,time2;

    int index=;

    for(int i=;i<=n;i++){

        if(h[i]-p[index]>x) return ;

        if(p[index]>=h[i]){//直接往右扫描

            while(index<=m && p[index]<=x+h[i])

                index++;

            if(index>m) return ;//

        }

        else {

            time1=(x-(h[i]-p[index]))/;

            time2=x-(h[i]-p[index])*;

            time1=max(time1,time2); //先往左在往右或者先往右再往左的最长右走时间

            while(index<=m && p[index]<=time1+h[i])

                index++;

            if(index>m) return ;

        }

    }

    return ;

}

int main(){

    while(scanf("%d%d",&n,&m)==){

        for(int i=;i<=n;i++) scanf("%lld",&h[i]);

        for(int i=;i<=m;i++) scanf("%lld",&p[i]);

        ll l=,r=max(h[n],p[m])*,ans=;

        while(l<=r){

            ll mid=l+r>>;

            if(judge(mid))

                ans=mid,r=mid-;

            else l=mid+;

        }

        printf("%I64d\n",ans);

    }

}