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;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int n, T[4 * 100005], T1[4 * 100005], p[100005], cep[100005], sag[100005];
int a[100005], mx, pos;
stack<pair<int, int> > s;
int tap1(int a, int b, int l, int r, int nd) {
  if (l > b or r < a) return -1000000009;
  if (l >= a and r <= b) return T1[nd];
  return max(tap1(a, b, l, (l + r) / 2, nd * 2),
             tap1(a, b, (l + r) / 2 + 1, r, nd * 2 + 1));
}
int tap(int a, int b, int l, int r, int nd) {
  if (l > b or r < a) return 1000000009;
  if (l >= a and r <= b) return T[nd];
  return min(tap(a, b, l, (l + r) / 2, nd * 2),
             tap(a, b, (l + r) / 2 + 1, r, nd * 2 + 1));
}
void build(int l, int r, int nd) {
  if (l == r) {
    T[nd] = p[l];
    T1[nd] = p[l];
    return;
  }
  build(l, (l + r) / 2, nd * 2);
  build((l + r) / 2 + 1, r, nd * 2 + 1);
  T[nd] = min(T[nd * 2], T[nd * 2 + 1]);
  T1[nd] = max(T1[nd * 2], T1[nd * 2 + 1]);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    while (!s.empty() and s.top().first <= a[i]) s.pop();
    if (s.empty())
      cep[i] = -1;
    else
      cep[i] = s.top().second;
    s.push({a[i], i});
    p[i] = p[i - 1] + a[i];
  }
  while (!s.empty()) s.pop();
  for (int i = n; i >= 1; i--) {
    while (!s.empty() and s.top().first <= a[i]) s.pop();
    if (s.empty())
      sag[i] = -1;
    else
      sag[i] = s.top().second;
    s.push({a[i], i});
  }
  build(0, n + 1, 1);
  for (int i = 1; i <= n; i++) {
    int x = 0, y = 0;
    if (cep[i] == -1)
      pos = 0;
    else
      pos = cep[i];
    x = p[i - 1] - tap(pos, i - 1, 0, n + 1, 1);
    if (sag[i] == -1)
      pos = n;
    else
      pos = sag[i] - 1;
    y = tap1(i, pos, 0, n + 1, 1) - p[i];
    mx = max(x + y, mx);
  }
  printf("%d", mx);
}
