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>
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int random(int l, int r) { return uniform_int_distribution<int>(l, r)(rng); }
int log2i(int x) { return x ? 31 - __builtin_clz(x) : -1; }
const int N = 1e5 + 2;
struct T {
  long long m, l, r, s;
  T() {}
  T(long long x) {
    s = x;
    m = l = r = max(x, 0LL);
  }
};
T operator+(T a, T b) {
  T ans;
  ans.l = max(a.l, a.s + b.l);
  ans.r = max(b.r, b.s + a.r);
  ans.s = a.s + b.s;
  ans.m = max(a.m, b.m);
  ans.m = max(ans.m, a.r + b.l);
  return ans;
}
T s[N];
int p[N];
int Find(int x) { return x == p[x] ? x : p[x] = Find(p[x]); }
long long ans = 0;
void Union(int x, int y) {
  x = Find(x);
  y = Find(y);
  if (x == y) return;
  if (x > y) swap(x, y);
  s[x] = s[x] + s[y];
  ans = max(ans, s[x].m);
  p[y] = x;
}
vector<int> pos[62];
bool vis[N];
void run() {
  int n, x;
  cin >> n;
  for (int i = 0; i < n; i++)
    cin >> x, s[i] = T(x), p[i] = i, pos[x + 30].push_back(i);
  long long res = 0;
  for (int i = 0; i < 62; i++) {
    for (auto u : pos[i]) {
      vis[u] = true;
      if (u + 1 < n && vis[u + 1]) Union(u, u + 1);
      if (u - 1 >= 0 && vis[u - 1]) Union(u - 1, u);
    }
    if (i >= 30) res = max(res, ans - i + 30);
  }
  cout << res << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  for (int i = 0; i < t; i++) {
    run();
  }
}
