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;
const int N = (int)2e5 + 123, inf = 1e9, mod = 1e9 + 7;
const long long INF = 1e18;
int n, pref[N], suf[N], sum[N], p[N], ans;
bool was[N];
pair<int, int> a[N];
int pp(int i) { return (p[i] == i ? i : p[i] = pp(p[i])); }
void Merge(int i, int j) {
  i = pp(i);
  j = pp(j);
  p[j] = i;
  pref[i] = max(pref[i], sum[i] + pref[j]);
  suf[i] = max(suf[j], sum[j] + suf[i]);
  sum[i] = sum[i] + sum[j];
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i].first);
    a[i].second = i;
  }
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    int j = a[i].second;
    was[j] = 1;
    p[j] = j;
    pref[j] = max(pref[j], a[i].first);
    suf[j] = max(suf[j], a[i].first);
    sum[j] = a[i].first;
    int cur = 0;
    if (j + 1 < n && was[j + 1]) cur += pref[pp(j + 1)];
    if (j - 1 >= 0 && was[j - 1]) cur += suf[pp(j - 1)];
    ans = max(ans, cur);
    if (j + 1 < n && was[j + 1]) Merge(j, j + 1);
    if (j - 1 >= 0 && was[j - 1]) Merge(j - 1, j);
  }
  printf("%d", ans);
}
