#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
using namespace std;
#define ll long long
#define lc u<<1
#define rc u<<1|1
#define X first
#define Y second
const int N = 1e5 + 5;
const int M = 5005;
const ll maxm = 1e18 + 5;

struct node {
    ll pos, state;
};
vector<node>mp;
ll ans = 0;

int cmp(node a, node b) {
    if (a.pos < b.pos)return 1;
    return 0;
}

ll R = 1e9 + 1;

void solve()
{
    //再战ant，思路没有问题，先按周期走，然后
    //懂了，还是距离的问题，越是到最后，越不能急，写的时候急
    //写完了，就应该仔细考究
    //模拟过程并不可取
    //仅在一次循环的时候才需要模拟
    ll n, l, r; cin >> n >> l >> r;
    //一个周期，我认为是左右，都减
    ll ans = 0;
    ll round = (min(l, r) / n);//当一端出现缺口，我们进行两次大循环模拟即可
    //我们认为一次大循环指的是所有的向右走的都撞了右壁，所有的向左走的都撞了左壁
    //一轮round指的是所有节点左撞一次右撞一次回到原点的过程
    ans += R * round * 2;
    l -= round * n; r -= round * n;
    //先让左走全撞，右走全撞
    //我缺乏快速切题的能力，我是说难题
    for (int i = 0; i < n; i++) {
        ll val; cin >> val;
        mp.push_back({ val,i });
    }
    for (int i = 0; i < n; i++) {
        cin >> mp[i].state;
    }
    //我明明就想要继续写，然后上蓝，然后挂头像，继续

    sort(mp.begin(), mp.end(), cmp);
    ll mx = 0;//记录移动的最大步数，障碍物其实是放置在0和1e9+1的位置上，如果有墙就返回，无走到那里就判定为掉下去
    for (int i = 0; i < n; i++) {
        //if (mp[i].state != 0)break;//让所有左走的都撞
        if (mp[i].state == 0) {
            if (l) {
                l--;
                mx = max(mp[i].pos, mx);
            }
            else {
                mp[i].state = 77;
                mx = max(mp[i].pos, mx);
            }
        }
    }
    for (int i = 0; i < n; i++) {
        if (mp[i].state == 1) {
            if (r) {
                r--;
                mx = max(R - mp[i].pos, mx);
            }
            else {
                mp[i].state = 77;
                mx = max(R - mp[i].pos, mx);
            }
        }
    }
    ans += mx;
    for (int i = 0; i < n; i++) {
        if (mp[i].state == 0) {
            mp[i].state = 1;
            //左走先走到0然后右走
            mp[i].pos = (mx - mp[i].pos);
        }
        else if (mp[i].state == 1) {
            mp[i].state = 0;
            mp[i].pos = R - (mx - (R - mp[i].pos));
        }
    }
    //更新状态再走一次
    /*
    cout << ans << endl;
    for (int i = 0; i < n; i++) {
        cout << mp[i].pos << " " << mp[i].state << endl;
    }*/


    sort(mp.begin(), mp.end(), cmp);
    mx = 0;//记录移动的最大步数，障碍物其实是放置在0和1e9+1的位置上，如果有墙就返回，无走到那里就判定为掉下去
    for (int i = 0; i < n; i++) {
        //if (mp[i].state != 0)break;//让所有左走的都撞
        if (mp[i].state == 0) {
            if (l) {
                l--;
                mx = max(mp[i].pos, mx);
            }
            else {
                mp[i].state = 77;
                mx = max(mp[i].pos, mx);
            }
        }
    }
    for (int i = n-1; i>=0; i--) {
        if (mp[i].state == 1) {
            if (r) {
                r--;
                mx = max(R - mp[i].pos, mx);
            }
            else {
                mp[i].state = 77;
                mx = max(R - mp[i].pos, mx);
            }
        }
    }
    ans += mx;
    for (int i = 0; i < n; i++) {
        if (mp[i].state == 0) {
            mp[i].state = 1;
            //左走先走到0然后右走
            mp[i].pos = (mx - mp[i].pos);
        }
        else if (mp[i].state == 1) {
            mp[i].state = 0;
            mp[i].pos = R - (mx - (R - mp[i].pos));
        }
    }
    //更新状态再走一次
    /*
    cout << ans << endl;
    for (int i = 0; i < n; i++) {
        cout << mp[i].pos << " " << mp[i].state << endl;
    }*/

    sort(mp.begin(), mp.end(), cmp);
    mx = 0;//记录移动的最大步数，障碍物其实是放置在0和1e9+1的位置上，如果有墙就返回，无走到那里就判定为掉下去
    for (int i = 0; i < n; i++) {
        //if (mp[i].state != 0)break;//让所有左走的都撞
        if (mp[i].state == 0) {
            if (l) {
                l--;
                mx = max(mp[i].pos, mx);
            }
            else {
                mp[i].state = 77;
                mx = max(mp[i].pos, mx);
            }
        }
    }
    for (int i = n-1; i>=0; i--) {
        if (mp[i].state == 1) {
            if (r) {
                r--;
                mx = max(R - mp[i].pos, mx);
            }
            else {
                mp[i].state = 77;
                mx = max(R - mp[i].pos, mx);
            }
        }
    }
    ans += mx;
    for (int i = 0; i < n; i++) {
        if (mp[i].state == 0) {
            mp[i].state = 1;
            //左走先走到0然后右走
            mp[i].pos = (mx - mp[i].pos);
        }
        else if (mp[i].state == 1) {
            mp[i].state = 0;
            mp[i].pos = R - (mx - (R - mp[i].pos));
        }
    }
    //更新状态再走一次
    //cout << ans << endl;

    sort(mp.begin(), mp.end(), cmp);
    mx = 0;
    for (int i = 0; i < n; i++) {
        //if (mp[i].state != 0)break;//让所有左走的都撞
        if (mp[i].state == 0) {
            if (l) {
                l--;
                mx = max(mp[i].pos, mx);
            }
            else {
                mp[i].state = 77;
                mx = max(mp[i].pos, mx);
            }
        }
    }
    for (int i = 0; i>=0; i--) {
        if (mp[i].state == 1) {
            if (r) {
                r--;
                mx = max(R - mp[i].pos, mx);
            }
            else {
                mp[i].state = 77;
                mx = max(R - mp[i].pos, mx);
            }
        }
    }
    ans += mx;
    for (int i = 0; i < n; i++) {
        if (mp[i].state == 0) {
            mp[i].state = 1;
            //左走先走到0然后右走
            mp[i].pos = (mx - mp[i].pos);
        }
        else if (mp[i].state == 1) {
            mp[i].state = 0;
            mp[i].pos = R - (mx - (R - mp[i].pos));
        }
    }
    //更新状态再走一次


    cout << ans << endl;
    return;
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int T = 1;
    //std::cin >> T;
    while (T--)
        solve();
    return 0;
}

ll ksm(ll a, ll b) {
    ll base = a;
    ll ans = 1;
    while (b) {
        if (b & 1)ans *= base;
        base *= base;
        b >>= 1;
    }
    return ans;
}
ll gcd(ll a, ll b) {
    return b ? gcd(b, a % b) : a;
}
/*致我已死的友人*/
/*PonyHex*/