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;
int N;
int A[101010];
int S[101010];
template <class V, int NV>
class SegTree_1 {
 public:
  vector<V> val;
  static V const def = -(1 << 30);
  V comp(V l, V r) { return max(l, r); };
  SegTree_1() { val = vector<V>(NV * 2, def); };
  V getval(int x, int y, int l = 0, int r = NV, int k = 1) {
    if (r <= x || y <= l) return def;
    if (x <= l && r <= y) return val[k];
    return comp(getval(x, y, l, (l + r) / 2, k * 2),
                getval(x, y, (l + r) / 2, r, k * 2 + 1));
  }
  void update(int entry, V v) {
    entry += NV;
    val[entry] = comp(v, val[entry]);
    while (entry > 1)
      entry >>= 1, val[entry] = comp(val[entry * 2], val[entry * 2 + 1]);
  }
};
SegTree_1<int, 1 << 20> st, st2;
vector<pair<int, int>> C;
void solve() {
  int i, j, k, l, r, x, y;
  string s;
  cin >> N;
  for (i = 0; i < (N); i++) {
    cin >> A[i];
    C.push_back({A[i], i});
    S[i + 1] = S[i] + A[i];
  }
  for (i = 0; i < (N + 1); i++) {
    st.update(i, S[i]);
    st2.update(i, -S[i]);
  }
  set<int> did;
  did.insert(-1);
  did.insert(N);
  sort((C.begin()), (C.end()));
  reverse((C.begin()), (C.end()));
  int ret = 0;
  for (auto& c : C) {
    x = c.second;
    auto it = did.lower_bound(x);
    int R = *it;
    int L = *prev(it);
    int ma = st.getval(x + 1, R + 1) + st2.getval(L + 1, x + 1) - A[x];
    ret = max(ret, ma);
    did.insert(x);
  }
  cout << ret << endl;
}
int main(int argc, char** argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
  for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
  cout.tie(0);
  solve();
  return 0;
}
