D. Yet Another Yet Another Task （區間最大和）

D. Yet Another Yet Another Task

Alice and Bob are playing yet another card game. This time the rules are the following. There are nn cards lying in a row in front of them. The ii-th card has value aiai.

First, Alice chooses a non-empty consecutive segment of cards [l;r][l;r] (l≤rl≤r). After that Bob removes a single card jj from that segment (l≤j≤r)(l≤j≤r). The score of the game is the total value of the remaining cards on the segment(al+al+1+⋯+aj−1+aj+1+⋯+ar−1+ar)(al+al+1+⋯+aj−1+aj+1+⋯+ar−1+ar). In particular, if Alice chooses a segment with just one element, then the score after Bob removes the only card is 00.

Alice wants to make the score as big as possible. Bob takes such a card that the score is as small as possible.

What segment should Alice choose so that the score is maximum possible? Output the maximum score.

Input

The first line contains a single integer nn (1≤n≤1051≤n≤105) — the number of cards.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (−30≤ai≤30−30≤ai≤30) — the values on the cards.

Output

Print a single integer — the final score of the game.

Examples

input

Copy

5
5 -2 10 -1 4

output

Copy

6

input

Copy

8
5 2 5 3 -30 -30 6 9

output

Copy

10

input

Copy

3
-10 6 -15

output

Copy

0

Note

In the first example Alice chooses a segment [1;5][1;5] — the entire row of cards. Bob removes card 33 with the value 1010 from the segment. Thus, the final score is 5+(−2)+(−1)+4=65+(−2)+(−1)+4=6.

In the second example Alice chooses a segment [1;4][1;4], so that Bob removes either card 11 or 33 with the value 55, making the answer 5+2+3=105+2+3=10.

In the third example Alice can choose any of the segments of length 11: [1;1][1;1], [2;2][2;2] or [3;3][3;3]. Bob removes the only card, so the score is 00. If Alice chooses some other segment then the answer will be less than 00.

 

#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <list>
#include <deque>
#include <queue>
#include <stack>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <iomanip>
#define pi acos(-1)
#define ull unsigned long long
#define ll long long
#define pb push_back
#define all(vc) vc.begin() , vc.end()
#define rep(i,start,end) for(int i=start;i<=end;i++)
#define per(i,end,start) for(int i=end;i>=start;i--)
#define tle ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define lc now<<1
#define rc now<<1|1
ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
using namespace std;
const int mod = 998244353;
const int mxn = 1e6 +7;
const int inf = 1e9;
int _,n,m,t,k,u,v,w,ans,cnt,ok,lim,len,tmp,last,nx;
struct node {int u,v,nx,w;}e[mxn];
string str;
int a[mxn];
void solve()
{
    ans = 0 ;
    for(int i=1;i<=30;i++)
    {
        int nowans = 0 , nowmin = 0 ;
        for(int j=1;j<=n;j++)
        {
            if(a[j]>i){
                nowans = 0 , nowmin = 0 ; continue;
            }
            nowans += a[j] ;
            nowmin = min( nowmin , nowans  );
            ans = max(nowans-nowmin-i , ans);
        }
    }
    cout<<ans<<endl;
}
int main()
{
    tle;
    while(cin>>n)
    {
        for(int i=1;i<=n;i++) cin>>a[i];
        solve();
    }
}