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 long long N = 111111;
long long n, a[N], pref[N], t[N][30], lg[N], ind[63];
long long gmin(long long l, long long r) {
  if (l > r) return 1e9;
  long long len = lg[r - l + 1];
  return min(t[l][len], t[r - (1ll << len) + 1][len]);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cout.precision(40);
  cin >> n;
  fill(ind, ind + 63, -1);
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
  }
  pref[0] = a[0];
  t[0][0] = a[0];
  for (long long i = 1; i < n; ++i) {
    pref[i] = pref[i - 1] + a[i];
    t[i][0] = pref[i];
  }
  for (long long k = 1; k < 20; k++) {
    long long len = 1ll << k;
    for (long long i = 0; i + len - 1 < n; ++i) {
      t[i][k] = min(t[i][k - 1], t[i + len / 2][k - 1]);
    }
  }
  for (long long i = 0; i < N; ++i) {
    long long j = 0;
    while ((1ll << j) <= i) j++;
    lg[i] = j - 1;
  }
  long long res = 0;
  for (long long i = 0; i < n; ++i) {
    long long mxind = 0;
    ind[a[i] + 30] = i;
    for (long long x = 62; x >= a[i] + 30; --x) {
      res = max(res, pref[i] - gmin(mxind, i - 1) - (x - 30));
      res = max(res, pref[i] - (mxind > 0 ? pref[mxind - 1] : 0) - (x - 30));
      mxind = max(mxind, ind[x] + 1);
    }
  }
  cout << res;
}
