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.
#include <bits/stdc++.h>
using namespace std;
int ans, A[100000];
void solve(int s, int e) {
  if (s == e) return;
  int m = (s + e) >> 1, LM[61], RM[61], mx = -31, sum = 0;
  solve(s, m);
  solve(m + 1, e);
  memset(LM, 0xC0, sizeof(LM));
  memset(RM, 0xC0, sizeof(RM));
  for (int i = m; i >= s; i--) {
    mx = max(mx, A[i]);
    sum += A[i];
    LM[mx + 30] = max(LM[mx + 30], sum);
  }
  sum = 0;
  mx = -31;
  for (int i = m + 1; i <= e; i++) {
    mx = max(mx, A[i]);
    sum += A[i];
    RM[mx + 30] = max(RM[mx + 30], sum);
  }
  for (int i = 0; i <= 60; i++)
    for (int j = 0; j <= 60; j++)
      ans = max(ans, LM[i] + RM[j] - (max(i, j) - 30));
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ((void)0);
  ((void)0);
  ((void)0);
  int N;
  cin >> N;
  for (int i = 0; i < N; i++) cin >> A[i];
  solve(0, N - 1);
  cout << ans << '\n';
  return 0;
}
