#include<stdio.h>

 #include<algorithm>

 using namespace std;

 const int N = ;

 long long a[N];

 int main(){

     int n, i;

     long long ma_ed, ans, sum = ;

     scanf("%d", &n);

     for(i = ; i <= n; ++i){

         scanf("%I64d", &a[i]);

         sum += a[i];

     }

     ma_ed = a[];

     ans = max(0LL, a[]);

     for(i = ; i <= n; ++i){//1~n求最大子段和

         if(ma_ed > )

             ma_ed += a[i];

         else

             ma_ed = a[i];

         ans = max(ma_ed, ans);

     }

     long long mi_ed = a[], s = a[];

     //最大字段和首尾相接情况

     for(i = ; i <= n; ++i){

         if(mi_ed < )

             mi_ed += a[i];

         else

             mi_ed = a[i];

         s = min(mi_ed, s);

     }

     //首尾相接时，答案为序列数的总和 与 其中数相加最小的和（负值） 之差

     printf("%I64d\n", max(ans, sum - s));

     return ;

 }