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;
long long n, a[100005], it[100005][100], it2[100005][100], sum[100005],
    it3[1000];
void buil() {
  for (long long i = (1); i <= (n); ++i) it[i][0] = a[i];
  for (long long k = 1; (1ll << k) <= n; k++)
    for (long long i = 1; i + (1ll << k) - 1 <= n; i++)
      it[i][k] = max(it[i][k - 1], it[i + (1 << (k - 1))][k - 1]);
}
long long get(long long l, long long r) {
  long long tmp = log2(r - l + 1);
  return max(it[l][tmp], it[r - (1ll << tmp) + 1][tmp]);
}
void buil2() {
  for (long long i = (1); i <= (n); ++i) it2[i][0] = sum[i];
  for (long long k = 1; (1ll << k) <= n; k++)
    for (long long i = 1; i + (1ll << k) - 1 <= n; i++)
      it2[i][k] = min(it2[i][k - 1], it2[i + (1 << (k - 1))][k - 1]);
}
long long get2(long long l, long long r) {
  if (l > r) return 1000000000;
  long long tmp = log2(r - l + 1);
  return min(it2[l][tmp], it2[r - (1ll << tmp) + 1][tmp]);
}
void up(long long k, long long l, long long r, long long i, long long val) {
  if (l > r) return;
  if (i < l || i > r) return;
  if (l == r && i == l) {
    it3[k] = val;
    return;
  }
  long long mid = (l + r) / 2;
  up(k * 2, l, mid, i, val);
  up(k * 2 + 1, mid + 1, r, i, val);
  it3[k] = max(it3[k * 2], it3[k * 2 + 1]);
}
long long get3(long long k, long long l, long long r, long long u,
               long long v) {
  if (l > r || v < l || u > r) return 0;
  if (l >= u && r <= v) return it3[k];
  long long mid = (l + r) / 2;
  return max(get3(k * 2, l, mid, u, v), get3(k * 2 + 1, mid + 1, r, u, v));
}
int main() {
  cin >> n;
  for (long long i = (1); i <= (n); ++i) {
    cin >> a[i];
    sum[i] = sum[i - 1] + a[i];
  }
  buil();
  buil2();
  long long ans = 0;
  for (long long i = (1); i <= (n); ++i) {
    up(1, 1, 30 + 31, a[i] + 31, i);
    for (long long k = (-30); k <= (30); ++k) {
      long long l = get3(1, 1, 30 + 31, k + 1 + 31, 30 + 31);
      if (l == 0)
        l = 1;
      else
        l++;
      if (l > i) continue;
      ans = max(ans, sum[i] - k - get2(l, i - 1));
      ans = max(ans, sum[i] - k - sum[l - 1]);
    }
  }
  cout << ans;
  return 0;
}
