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

First, Alice chooses a non-empty consecutive segment of cards [l; r] (l ≤ r). After that Bob removes a single card j from that segment (l ≤ j ≤ r). The score of the game is the total value of the remaining cards on the segment (a_l + a_{l + 1} + ... + a_{j - 1} + a_{j + 1} + ... + a_{r - 1} + a_r). In particular, if Alice chooses a segment with just one element, then the score after Bob removes the only card is 0.

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 n (1 ≤ n ≤ 10^5) — the number of cards.

The second line contains n integers a_1, a_2, ..., a_n (-30 ≤ a_i ≤ 30) — the values on the cards.

Output

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

Examples

Input


5
5 -2 10 -1 4


Output


6


Input


8
5 2 5 3 -30 -30 6 9


Output


10


Input


3
-10 6 -15


Output


0

Note

In the first example Alice chooses a segment [1;5] — the entire row of cards. Bob removes card 3 with the value 10 from the segment. Thus, the final score is 5 + (-2) + (-1) + 4 = 6.

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

In the third example Alice can choose any of the segments of length 1: [1;1], [2;2] or [3;3]. Bob removes the only card, so the score is 0. If Alice chooses some other segment then the answer will be less than 0.
import sys
input = sys.stdin.readline
from itertools import accumulate

n=int(input())
A=list(map(int,input().split()))

S=[0]+list(accumulate(A))

import heapq

XMIN=[[] for i in range(61)]
XMAX=[[] for i in range(61)]

for i in range(n):
    XMIN[A[i]+30].append(i)
    XMAX[A[i]+30].append(-i)

for i in range(61):
    heapq.heapify(XMIN[i])
    heapq.heapify(XMAX[i])

seg_el=1<<((n+2).bit_length())
SEGMIN=[1<<30]*(2*seg_el)
SEGMAX=[-1<<30]*(2*seg_el)

for i in range(n+1):
    SEGMIN[i+seg_el]=S[i]
    SEGMAX[i+seg_el]=S[i]

for i in range(seg_el-1,0,-1):
    SEGMIN[i]=min(SEGMIN[i*2],SEGMIN[i*2+1])
    SEGMAX[i]=max(SEGMAX[i*2],SEGMAX[i*2+1])

def update(n):
    i=n+seg_el
    SEGMIN[i]=1<<30
    SEGMAX[i]=-1<<30
    i>>=1
    
    while i!=0:
        SEGMIN[i]=min(SEGMIN[i*2],SEGMIN[i*2+1])
        SEGMAX[i]=max(SEGMAX[i*2],SEGMAX[i*2+1])
        i>>=1
 
def getvalues_min(l,r):
    L=l+seg_el
    R=r+seg_el
    ANS=1<<30

    while L<R:
        if L & 1:
            ANS=min(ANS , SEGMIN[L])
            L+=1

        if R & 1:
            R-=1
            ANS=min(ANS , SEGMIN[R])
        L>>=1
        R>>=1

    return ANS

def getvalues_max(l,r):
    L=l+seg_el
    R=r+seg_el
    ANS=-1<<30

    while L<R:
        if L & 1:
            ANS=max(ANS , SEGMAX[L])
            L+=1

        if R & 1:
            R-=1
            ANS=max(ANS , SEGMAX[R])
        L>>=1
        R>>=1

    return ANS

A1=[0]*n
A2=[0]*n
for i in range(n):
    v=A[i]+30

    MIN=n+1

    for j in range(v+1,61):
        while XMIN[j] and XMIN[j][0]<i:
            heapq.heappop(XMIN[j])
        if XMIN[j]:
            MIN=min(MIN,XMIN[j][0])

    A1[i]=max(0,getvalues_max(i+2,MIN+1)-S[i+1])
    #print(i,MIN,getvalues_max(i+2,MIN+1),S[i+1])

for i in range(n-1,-1,-1):
    v=A[i]+30

    MAX=-1

    for j in range(v+1,61):
        while XMAX[j] and -XMAX[j][0]>i:
            heapq.heappop(XMAX[j])
        if XMAX[j]:
            MAX=max(MAX,-XMAX[j][0])

    A2[i]=max(0,S[i]-getvalues_min(MAX+1,i+1))
    #print(i,MAX,getvalues_min(MAX+1,i+1),S[i])

ANS=0
for i in range(n):
    ANS=max(ANS,A1[i]+A2[i])

print(ANS)

