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 inf = 0x3f3f3f3f;
const long long INF = 2e18;
const int OFFSET = 30;
const int VAL = 65;
int pos[VAL];
const int N = 100000 + 7;
int mx[4 * N], mn[4 * N];
void update(int at, int l, int r, int p, int x) {
  if (l == r) {
    mx[at] = x;
    mn[at] = x;
    return;
  }
  int lc = at + at, rc = lc + 1, mid = (l + r) / 2;
  if (p <= mid)
    update(lc, l, mid, p, x);
  else
    update(rc, mid + 1, r, p, x);
  mx[at] = max(mx[lc], mx[rc]);
  mn[at] = min(mn[lc], mn[rc]);
}
int query_min(int at, int l, int r, int lo, int hi) {
  if (l > hi or r < lo) return inf;
  if (l >= lo and r <= hi) return mn[at];
  int lc = at + at, rc = lc + 1, mid = (l + r) / 2;
  return min(query_min(lc, l, mid, lo, hi), query_min(rc, mid + 1, r, lo, hi));
}
int query_max(int at, int l, int r, int lo, int hi) {
  if (l > hi or r < lo) return -inf;
  if (l >= lo and r <= hi) return mx[at];
  int lc = at + at, rc = lc + 1, mid = (l + r) / 2;
  return max(query_max(lc, l, mid, lo, hi), query_max(rc, mid + 1, r, lo, hi));
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  vector<int> a(n);
  for (auto &qq : a) cin >> qq;
  vector<int> cum(n + 1, 0);
  for (int i = 1; i <= n; ++i) cum[i] = cum[i - 1] + a[i - 1];
  for (int i = 0; i <= n; ++i) update(1, 0, n, i, cum[i]);
  vector<int> res(n, 0);
  for (int i = 0; i < VAL; ++i) pos[i] = -1;
  for (int i = 0; i < n; ++i) {
    int cur = a[i] + OFFSET;
    int pre = -1;
    for (int j = cur + 1; j < VAL; ++j) pre = max(pre, pos[j]);
    res[i] += cum[i] - query_min(1, 0, n, pre + 1, i);
    pos[cur] = i;
  }
  for (int i = 0; i < VAL; ++i) pos[i] = n;
  for (int i = n - 1; i >= 0; --i) {
    int cur = a[i] + OFFSET;
    int pre = n;
    for (int j = cur + 1; j < VAL; ++j) pre = min(pre, pos[j]);
    res[i] += query_max(1, 0, n, i + 1, pre) - cum[i + 1];
    pos[cur] = i;
  }
  int ret = *max_element((res).begin(), (res).end());
  cout << ret << "\n";
  return 0;
}
