#pragma comment(linker, "/STACK:65777216")
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <string>
#include <string.h>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <time.h>
#include <cmath>
#include <algorithm>
#include <malloc.h>

#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define sz(x) ((int) (x).size())
#define forn(var, lim) for(int var=0; var<lim; ++var)
#define all(x) (x).begin(), (x).end()
#define abs(x) ((x)>=0 ? (x) : -(x))


using namespace std;

typedef long long ll;
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<bool> vb;
typedef vector<ii> vii;
typedef vector<vi> vvi;
typedef vector<vb> vvb;



struct she{
    int x, num;
};

struct team{
    int x, num, sh;
};

bool inline operator < (she a, she b) {
    return a.x<b.x;
}

bool inline operator < (team a, team b) {
    return a.x<b.x;
}

int main(int argc, char **argv) {
    int n, m;

    scanf("%d", &n);
    team* b=new team[n];
    forn(i, n) {
        scanf("%d", &b[i].x);
        b[i].num=i;
    }
    sort(b, b+n);

    scanf("%d", &m);
    she* a=new she[m];
    forn(i, m) {
        scanf("%d", &a[i].x);
        a[i].num=i;
    }
    sort(a, a+m);
    vvi c(n);

    forn(i, n) {
        b[i].sh=a[i].num;
        c[i].pb(i);
    }
    for(int i=m; i<n; i++) {
        b[i].sh=a[m-1].num;
        ll cmax=0;
        int ind;
        ll ct=0;
        for(int k=m-1; k>0 && c[k].size()==1;k--) {
            ct+=abs(b[c[k][0]].x-a[k].x)-abs(b[c[k][0]].x-a[k-1].x);
            if(ct>cmax) {
                cmax=ct;
                ind=k;
            }
        }
        if(cmax>0) {
            for(int k=ind; k<m; k++) {
                b[c[k][0]].sh=a[k-1].num;
                c[k-1].pb(c[k][0]);
                c[k].clear();
            }
        }
        c[m-1].pb(i);
    }
    ll ans=0;
    vi res(n);
    forn(i, m) {
        forn(j, c[i].size()) {
            ans+=abs(b[c[i][j]].x-a[i].x);
            res[b[c[i][j]].num]=a[i].num;
        }
    }
    cout << ans << endl;
    forn(i, n) {
        if(i) printf(" ");
        printf("%d", res[i]+1);
    }
    printf("\n");
    return 0;
}