The circle line of the Berland subway has *n* stations. We know the distances between all pairs of neighboring stations:
-  *d*<sub class="lower-index">1</sub> is the distance between the 1-st and the 2-nd station;-  *d*<sub class="lower-index">2</sub> is the distance between the 2-nd and the 3-rd station;...-  *d*<sub class="lower-index">*n* - 1</sub> is the distance between the *n* - 1-th and the *n*-th station;-  *d*<sub class="lower-index">*n*</sub> is the distance between the *n*-th and the 1-st station.
The trains go along the circle line in both directions. Find the shortest distance between stations with numbers *s* and *t*.
The first line contains integer *n* (3 ≤ *n* ≤ 100) — the number of stations on the circle line. The second line contains *n* integers *d*<sub class="lower-index">1</sub>, *d*<sub class="lower-index">2</sub>, ..., *d*<sub class="lower-index">*n*</sub> (1 ≤ *d*<sub class="lower-index">*i*</sub> ≤ 100) — the distances between pairs of neighboring stations. The third line contains two integers *s* and *t* (1 ≤ *s*, *t* ≤ *n*) — the numbers of stations, between which you need to find the shortest distance. These numbers can be the same.

The numbers in the lines are separated by single spaces.
#include <iostream>
using namespace std;

int main() {
    int n;
    cin>>n;
    int d[n];
    for (int i=0; i < n; i++){ cin>>d[i]; }
    int a,b;
    cin>>a>>b;
    int sum1 =0 , sum2 = 0;

    if(a==0){a=n-1;}
    else{a=a-1;}

    if(b==0){b=n-2;}
    else if(b==1){b=n-1;}
    else {b=b-2;}

    int i=a;
    while(i!=((b+1)%n)){
        sum1+=d[i];
        i = (i+1)%n;
    }
    i=(b+1)%n;

    while(i!=a){
        sum2+=d[i];
        i = (i+1)%n;
    }
    cout<<min(sum1,sum2)<<endl;
    return 0;
}