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;
vector<long long> mint, maxt;
vector<long long> dp;
void build(long long v, long long l, long long r) {
  if (l + 1 == r) {
    mint[v] = maxt[v] = dp[l];
    return;
  }
  long long m = (r + l) / 2;
  build(2 * v + 1, l, m);
  build(2 * v + 2, m, r);
  mint[v] = min(mint[2 * v + 1], mint[2 * v + 2]);
  maxt[v] = max(maxt[2 * v + 1], maxt[2 * v + 2]);
}
long long maxi(long long v, long long l, long long r, long long L,
               long long R) {
  if (l >= L && r <= R) {
    return maxt[v];
  }
  if (l >= R || r <= L) {
    return -1e18;
  }
  long long m = (r + l) / 2;
  return max(maxi(2 * v + 1, l, m, L, R), maxi(2 * v + 2, m, r, L, R));
}
long long mini(long long v, long long l, long long r, long long L,
               long long R) {
  if (l >= L && r <= R) {
    return mint[v];
  }
  if (l >= R || r <= L) {
    return 1e18;
  }
  long long m = (r + l) / 2;
  return min(mini(2 * v + 1, l, m, L, R), mini(2 * v + 2, m, r, L, R));
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  vector<long long> p(n);
  for (auto &x : p) {
    cin >> x;
  }
  dp.assign(n + 1, 0);
  for (int i = 0; i < n; i++) {
    dp[i + 1] = dp[i] + p[i];
  }
  mint.resize(4 * (dp.size()));
  maxt.resize(4 * (dp.size()));
  build(0, 0, dp.size());
  vector<map<long long, long long> > left(n + 1), right(n + 1);
  map<long long, long long> last;
  for (int i = -30; i <= 30; i++) {
    last[i] = -1;
  }
  for (int i = 0; i < n; i++) {
    left[i + 1] = last;
    last[p[i]] = i + 1;
  }
  for (int i = -30; i <= 30; i++) {
    last[i] = dp.size();
  }
  for (int i = n - 1; i >= 0; i--) {
    right[i + 1] = last;
    last[p[i]] = i + 1;
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    long long l = -1;
    for (int j = p[i] + 1; j <= 30; j++) {
      l = max(l, left[i + 1][j]);
    }
    long long r = dp.size();
    for (int j = p[i] + 1; j <= 30; j++) {
      r = min(r, right[i + 1][j]);
    }
    long long maxx = maxi(0, 0, dp.size(), i + 1, r);
    long long minn = mini(0, 0, dp.size(), l + 1, i + 1);
    ans = max(ans, maxx - minn - p[i]);
  }
  cout << ans;
  return 0;
}
