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;
inline long long fpow(long long n, long long k, long long p = 998244353) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
inline long long inv(long long a, long long p = 998244353) {
  return fpow(a, p - 2, p);
}
inline long long addmod(long long a, long long val, long long p = 998244353) {
  {
    if ((a = (a + val)) >= p) a -= p;
  }
  return a;
}
inline long long submod(long long a, long long val, long long p = 998244353) {
  {
    if ((a = (a - val)) < 0) a += p;
  }
  return a;
}
inline long long mult(long long a, long long b, long long p = 998244353) {
  return (long long)a * b % p;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  vector<long long> a(n);
  set<long long> s;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    s.insert(a[i]);
  }
  long long ans = 0;
  for (auto i : s) {
    vector<long long> pref(n + 1, 0);
    vector<long long> mark(n + 1, 0);
    bool check = false;
    long long maxi = LLONG_MIN;
    for (int j = 1; j <= n; j++) {
      if (a[j - 1] <= i) {
        pref[j] = max(a[j - 1], pref[j - 1] + a[j - 1]);
      }
      if (a[j - 1] == i) {
        check = true;
      } else if (a[j - 1] > i) {
        check = false;
        maxi = LLONG_MIN;
      }
      if (check) {
        maxi = max(maxi, pref[j]);
        ans = max(ans, maxi - i);
      }
    }
  }
  cout << ans << "\n";
}
