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;
bool isprime(long long int n) {
  long long int i, j;
  for (i = 2; i * i <= n; i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
long long int power(long long int x, long long int n, long long int mod) {
  if (x == 0) return 0;
  if (n == 0) return 1;
  if (n == 1) return x % mod;
  if (n % 2 == 0) {
    long long int temp = power(x, n / 2, mod);
    return (temp % mod * temp % mod) % mod;
  } else {
    long long int temp = power(x, n / 2, mod);
    return (x % mod * ((temp % mod * temp % mod) % mod) % mod) % mod;
  }
}
long long int modInverse(long long int a, long long int m) {
  return power(a, m - 2, m);
}
long long int findParent(long long int i, long long int parent[]) {
  if (parent[parent[i]] != parent[i]) parent[i] = findParent(parent[i], parent);
  return parent[i];
}
void unionNodes(long long int a, long long int b, long long int parent[],
                long long int size[]) {
  long long int parent_a = findParent(a, parent),
                parent_b = findParent(b, parent);
  if (parent_a == parent_b) return;
  if (size[parent_a] >= size[parent_b]) {
    size[parent_a] += size[parent_b];
    parent[parent_b] = parent_a;
  } else {
    size[parent_b] += size[parent_a];
    parent[parent_a] = parent_b;
  }
  return;
}
int main() {
  long long int n, i, j, k, temp;
  cin >> n;
  long long int arr[n];
  set<long long int> s;
  for (i = 0; i < n; i++) {
    cin >> arr[i];
    s.insert(arr[i]);
  }
  long long int ans = 0, sumi = 0;
  for (i = 1; i <= 30; i++) {
    temp = 0, sumi = 0;
    if (s.find(i) == s.end())
      continue;
    else {
      for (j = 0; j < n; j++) {
        if (arr[j] > i)
          sumi = 0;
        else
          sumi += arr[j];
        if (sumi < 0) sumi = 0;
        temp = max(temp, sumi);
      }
      ans = max(ans, temp - i);
    }
  }
  cout << ans << '\n';
  return 0;
}
