After battling Shikamaru, Tayuya decided that her flute is too predictable, and replaced it with a guitar. The guitar has 6 strings and an infinite number of frets numbered from 1. Fretting the fret number j on the i-th string produces the note a_{i} + j.

Tayuya wants to play a melody of n notes. Each note can be played on different string-fret combination. The easiness of performance depends on the difference between the maximal and the minimal indices of used frets. The less this difference is, the easier it is to perform the technique. Please determine the minimal possible difference.

For example, if a = [1, 1, 2, 2, 3, 3], and the sequence of notes is 4, 11, 11, 12, 12, 13, 13 (corresponding to the second example), we can play the first note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7, as shown on the picture.

<image>

Input

The first line contains 6 space-separated numbers a_{1}, a_{2}, ..., a_{6} (1 ≤ a_{i} ≤ 10^{9}) which describe the Tayuya's strings.

The second line contains the only integer n (1 ≤ n ≤ 100 000) standing for the number of notes in the melody.

The third line consists of n integers b_{1}, b_{2}, ..., b_{n} (1 ≤ b_{i} ≤ 10^{9}), separated by space. They describe the notes to be played. It's guaranteed that b_i > a_j for all 1≤ i≤ n and 1≤ j≤ 6, in other words, you can play each note on any string.

Output

Print the minimal possible difference of the maximal and the minimal indices of used frets.

Examples

Input


1 4 100 10 30 5
6
101 104 105 110 130 200


Output


0


Input


1 1 2 2 3 3
7
13 4 11 12 11 13 12


Output


7

Note

In the first sample test it is optimal to play the first note on the first string, the second note on the second string, the third note on the sixth string, the fourth note on the fourth string, the fifth note on the fifth string, and the sixth note on the third string. In this case the 100-th fret is used each time, so the difference is 100 - 100 = 0.

<image>

In the second test it's optimal, for example, to play the second note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7.

<image>
import java.util.*;
import java.io.*;
import java.text.*;

public class CF_1434_A{
    //SOLUTION BEGIN
    void pre() throws Exception{}
    void solve(int TC) throws Exception{
        int M = 6;
        long[] A = new long[M];
        for(int i = 0; i< M; i++)A[i] = nl();
        int N = ni();
        long[] B = new long[N];
        for(int i = 0; i< N; i++)B[i] = nl();

        long[][] pair = new long[M*N][];
        for(int i = 0; i< N; i++){
            for(int j = 0; j< M; j++){
                pair[i*M+j] = new long[]{B[i]-A[j], i};
            }
        }
        Arrays.sort(pair, (long[] l1, long[] l2) -> Long.compare(l1[0], l2[0]));
        int[] f = new int[N];
        int cnt = 0;
        long ans = (long)1e18;
        for(int i = 0, ptr = 0; i < pair.length; i++){
            while(ptr < pair.length && cnt < N){
                int ind = (int)pair[ptr++][1];
                if(f[ind] == 0)cnt++;
                f[ind]++;
            }
            if(cnt == N)ans = Math.min(ans, pair[ptr-1][0]-pair[i][0]);
            int ind = (int)pair[i][1];
            if(f[ind] == 1)cnt--;
            f[ind]--;
        }
        pn(ans);
    }
    //SOLUTION END
    void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
    void exit(boolean b){if(!b)System.exit(0);}
    static void dbg(Object... o){System.err.println(Arrays.deepToString(o));}
    final long IINF = (long)1e17;
    final int INF = (int)1e9+2;
    DecimalFormat df = new DecimalFormat("0.00000000000");
    double PI = 3.141592653589793238462643383279502884197169399, eps = 1e-8;
    static boolean multipleTC = false, memory = true, fileIO = false;
    FastReader in;PrintWriter out;
    void run() throws Exception{
        long ct = System.currentTimeMillis();
        if (fileIO) {
            in = new FastReader("");
            out = new PrintWriter("");
        } else {
            in = new FastReader();
            out = new PrintWriter(System.out);
        }
        //Solution Credits: Taranpreet Singh
        int T = multipleTC? ni():1;
        pre();
        for (int t = 1; t <= T; t++) solve(t);
        out.flush();
        out.close();
        System.err.println(System.currentTimeMillis() - ct);
    }
    public static void main(String[] args) throws Exception{
        if(memory)new Thread(null, new Runnable() {public void run(){try{new CF_1434_A().run();}catch(Exception e){e.printStackTrace();System.exit(1);}}}, "1", 1 << 28).start();
        else new CF_1434_A().run();
    }
    int[][] make(int n, int e, int[] from, int[] to, boolean f){
        int[][] g = new int[n][];int[]cnt = new int[n];
        for(int i = 0; i< e; i++){
            cnt[from[i]]++;
            if(f)cnt[to[i]]++;
        }
        for(int i = 0; i< n; i++)g[i] = new int[cnt[i]];
        for(int i = 0; i< e; i++){
            g[from[i]][--cnt[from[i]]] = to[i];
            if(f)g[to[i]][--cnt[to[i]]] = from[i];
        }
        return g;
    }
    int[][][] makeS(int n, int e, int[] from, int[] to, boolean f){
        int[][][] g = new int[n][][];int[]cnt = new int[n];
        for(int i = 0; i< e; i++){
            cnt[from[i]]++;
            if(f)cnt[to[i]]++;
        }
        for(int i = 0; i< n; i++)g[i] = new int[cnt[i]][];
        for(int i = 0; i< e; i++){
            g[from[i]][--cnt[from[i]]] = new int[]{to[i], i, 0};
            if(f)g[to[i]][--cnt[to[i]]] = new int[]{from[i], i, 1};
        }
        return g;
    }
    int find(int[] set, int u){return set[u] = (set[u] == u?u:find(set, set[u]));}
    int digit(long s){int ans = 0;while(s>0){s/=10;ans++;}return ans;}
    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
    void p(Object... o){for(Object oo:o)out.print(oo+" ");}
    void pn(Object... o){for(int i = 0; i< o.length; i++)out.print(o[i]+(i+1 < o.length?" ":"\n"));}
    void pni(Object... o){for(Object oo:o)out.print(oo+" ");out.println();out.flush();}
    String n()throws Exception{return in.next();}
    String nln()throws Exception{return in.nextLine();}
    int ni()throws Exception{return Integer.parseInt(in.next());}
    long nl()throws Exception{return Long.parseLong(in.next());}
    double nd()throws Exception{return Double.parseDouble(in.next());}

    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

        String next() throws Exception{
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    throw new Exception(e.toString());
                }
            }
            return st.nextToken();
        }

        String nextLine() throws Exception{
            String str;
            try{
                str = br.readLine();
            }catch (IOException e){
                throw new Exception(e.toString());
            }
            return str;
        }
    }
}