In distant future on Earth day lasts for n hours and that's why there are n timezones. Local times in adjacent timezones differ by one hour. For describing local time, hours numbers from 1 to n are used, i.e. there is no time "0 hours", instead of it "n hours" is used. When local time in the 1-st timezone is 1

 hour, local time in the i-th timezone is i hours.

Some online programming contests platform wants to conduct a contest that lasts for an hour in such a way that its beginning coincides with beginning of some hour (in all time zones). The platform knows, that there are ai people from i-th timezone who want to participate in the contest. Each person will participate if and only if the contest starts no earlier than s

 hours 00 minutes local time and ends not later than f hours 00 minutes local time. Values s and f are equal for all time zones. If the contest starts at f hours 00 minutes local time, the person won't participate in it.

Help platform select such an hour, that the number of people who will participate in the contest is maximum.

Input

The first line contains a single integer n (2 ≤ n ≤ 100 000) — the number of hours in day.

The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 10 000), where ai is the number of people in the i-th timezone who want to participate in the contest.

The third line contains two space-separated integers s and f (1 ≤ s < f ≤ n).

Output

Output a single integer — the time of the beginning of the contest (in the first timezone local time), such that the number of participants will be maximum possible. If there are many answers, output the smallest among them.

ExamplesinputCopy

3
1 2 3
1 3

output

3

inputCopy

5
1 2 3 4 1
1 3

output

4

Note

In the first example, it's optimal to start competition at 3 hours (in first timezone). In this case, it will be 1 hour in the second timezone and 2hours in the third timezone. Only one person from the first timezone won't participate.

In second example only people from the third and the fourth timezones will participate.

題意：在未來，一天一共有n個小時，也一共有n個時區，每兩個相鄰的時區之間差是一個小時。現在有一些網站想要舉辦線上程式設計比賽，希望有最多的人能參與。因為人們以自己的時區的時間為標準，只參加在s點以後（包括s點）開始的且在f點之前（包括f點）的比賽。每場比賽的時間為1個小時，讓你求出比賽在什麼時間開始會有最多的人蔘加，輸出以第一個時區的時間為標準的開始時間。

思路：因為比賽時間為1個小時，所以顯然能夠參加到比賽的時區一共有 (f - s) 個。這樣肯定就是求連續(f-s)個ai的最大是從哪個開始的。這題有一個坑點，就是第二天的1點是算是比當天的2點要早的，第二天的2點是比當天的1點要遲的，所以算連續的時候，如果後面不夠了，要補上前面的。還有就是要輸出最小的那個答案。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define LL long long

using namespace std;

LL sum[100010];

int main(void)
{
    int n,i,j;
    while(scanf("%d",&n)==1)
    {
        sum[0] = 0;
        for(i=1;i<=n;i++)
        {
            int x;
            scanf("%d",&x);
            sum[i] = sum[i-1] + x;
        }
        int s,f;
        scanf("%d%d",&s,&f);
        int t = f-s;
        LL maxn = 0;
        int ans = n;
        for(i=1;i<=n-t+1;i++)
        {
            LL tmp = sum[i+t-1] - sum[i-1];
            if(tmp > maxn)
            {
                maxn = tmp;
                ans = s - i + 1;
                if(ans <= 0)
                    ans += n;
            }
            else if(tmp == maxn)
            {
                int ta = s-i+1;
                if(ta <= 0)
                    ta += n;
                if(ta < ans)
                    ans = ta;
            }
        }
        for(;i<=n;i++)
        {
            LL tmp = sum[n] - sum[i-1] + sum[t-n+i-1];
            if(tmp > maxn)
            {
                maxn = tmp;
                ans = s - i + 1;
                if(ans <= 0)
                    ans += n;
            }
            else if(tmp == maxn)
            {
                int ta = s-i+1;
                if(ta <= 0)
                    ta += n;
                if(ta < ans)
                    ans = ta;
            }
        }

        printf("%d\n",ans);
    }


    return 0;
}