//POJ3377

//DP解法-解有规律的最短路问题

//Time:1157Ms   Memory:12440K

#include<iostream>

#include<cstring>

#include<cstdio>

using namespace std;

#define MAXN 1000005

typedef long long LL;

int n;

int dp[MAXN][3];

int sr, st, er, ed;

int main()

{

    //freopen("in.txt", "r", stdin);

    while(scanf("%d", &n), n)

    {

        scanf("%d%d%d%d", &sr,&st,&er,&ed);

        if(st > ed)

        {

            swap(st, ed);

            swap(sr, er);

        }

        for(int i = 1; i <= n; i++)

            scanf("%d", &dp[i][0]);

        for(int i = 0; i <= n; i++)

            scanf("%d", &dp[i][2]);

        for(int i = 1; i <= n; i++)

            scanf("%d", &dp[i][1]);

        //更新st从左侧到达对岸的最短路

        for(int i = st, w = 0; i > 0; i--)

        {

            w += dp[i][0] + dp[i][1];   //间接走陆路的路长和

            dp[st][2] = min(dp[st][2], w + dp[i-1][2]);

            if(w >= dp[st][2]) break;

        }

        //更新ed从右侧到达对岸的最短路

        for(int i = ed+1, w = 0; i <= n; i++)

        {

            w += dp[i][0] + dp[i][1];

            dp[ed][2] = min(dp[ed][2], w + dp[i][2]);

            if(w >= dp[ed][2])  break;

        }

        LL dis[2];

        dis[sr] = 0;    //起始点右移最短路

        dis[!sr] = dp[st][2];   //对岸右移最短路

        for(int i = st + 1; i <= ed; i++)

        {

            int x = dp[i][sr], y = dp[i][!sr];  //该点与对岸到达右一点的路长

            int z = dp[i][2];   //右侧水路长

            LL tmp = dis[sr];

            dis[sr] = min(dis[sr] + x, dis[!sr] + y + z);

            dis[!sr] = min(dis[!sr] + y, tmp + x + z);

        }

        printf("%lld\n", dis[er]);

    }

    return 0;

}