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 md = 1e9 + 7;
const long long MX = 2e5 + 5;
const long long INF = 1e18;
const long double PI = 4 * atan((long double)1);
long long power_md(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2) res = (res % md * a % md) % md;
    a = (a % md * a % md) % md;
    n /= 2;
  }
  res %= md;
  return res;
}
long long power(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2) res *= a;
    a = a * a;
    n /= 2;
  }
  return res;
}
long long abst(long long a) { return ((a < 0) ? (-1 * a) : (a)); }
double abstd(double x) {
  if (x < 0.0) x = -x;
  return x;
}
class cmp_set {
 public:
  bool operator()(long long a, long long b) { return a > b; }
};
vector<string> vec_splitter(string s) {
  s += ',';
  vector<string> res;
  while (!s.empty()) {
    res.push_back(s.substr(0, s.find(',')));
    s = s.substr(s.find(',') + 1);
  }
  return res;
}
void debug_out(vector<string> __attribute__((unused)) args,
               __attribute__((unused)) long long idx,
               __attribute__((unused)) long long LINE_NUM) {
  cerr << endl;
}
template <typename Head, typename... Tail>
void debug_out(vector<string> args, long long idx, long long LINE_NUM, Head H,
               Tail... T) {
  if (idx > 0)
    cerr << ", ";
  else
    cerr << "Line(" << LINE_NUM << ") ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, LINE_NUM, T...);
}
long long n;
long long arr[100005];
long long dp[100005][32][3];
long long fun(long long i, long long maxn, long long st) {
  if (i == n || (st == 2)) return (-maxn);
  if (dp[i][maxn][st] != -1) return dp[i][maxn][st];
  long long res = -1e9;
  long long tt;
  if (st == 0) {
    res = max(res, fun(i + 1, maxn, st));
    tt = max(maxn, arr[i]);
    res = max(res, fun(i + 1, tt, (long long)1) + arr[i]);
    res = max(res, fun(i + 1, tt, (long long)2) + arr[i]);
  } else if (st == 1) {
    tt = max(maxn, arr[i]);
    res = max(res, fun(i + 1, tt, (long long)1) + arr[i]);
    res = max(res, fun(i + 1, tt, (long long)2) + arr[i]);
  }
  return dp[i][maxn][st] = res;
}
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  memset(dp, -1, sizeof(dp));
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> arr[i];
  long long res = fun((long long)0, (long long)0, (long long)0);
  res = max(res, (long long)0);
  cout << res << "\n";
  return 0;
}
