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 isvowel(char c) {
  if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') return true;
  return false;
}
long long int isprime(long long int n) {
  for (long long int i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
long long int power(long long int a, long long int b) {
  if (b == 0) return 1;
  if (b == 1) return a % 998244353;
  long long int temp = power(a, b / 2);
  if (b % 2 == 0)
    return ((temp % 998244353) * (temp % 998244353)) % 998244353;
  else
    return ((((temp % 998244353) * (temp % 998244353)) % 998244353) * a %
            998244353) %
           998244353;
}
long long int c(long long int n, long long int k) {
  long long int C[n + 1][k + 1];
  long long int i, j;
  for (i = 0; i <= n; i++) {
    for (j = 0; j <= min(i, k); j++) {
      if (j == 0 || j == i)
        C[i][j] = 1;
      else
        C[i][j] = (C[i - 1][j - 1] % 998244353 + C[i - 1][j]) % 998244353;
    }
  }
  return C[n][k] % 998244353;
}
long long int modInverse(long long int a, long long int m) {
  long long int m0 = m;
  long long int y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long int q = a / m;
    long long int t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
vector<long long int> graph[200005];
long long int visited[200005] = {0};
long long int depth[200005];
long long int siz[200005];
void dfs(long long int i, long long int d) {
  visited[i] = 1;
  depth[i] = d;
  if (graph[i].size() == 1) siz[i] = 0;
  for (long long int j = 0; j < graph[i].size(); j++) {
    if (!visited[graph[i][j]]) {
      dfs(graph[i][j], d + 1);
      siz[i] += siz[graph[i][j]] + 1;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  t = 1;
  while (t--) {
    long long int n;
    cin >> n;
    long long int a[n];
    for (long long int i = 0; i < n; i++) cin >> a[i];
    long long int ans = 0;
    for (long long int i = 0; i < 31; i++) {
      long long int sum = 0, best = 0;
      for (long long int j = 0; j < n; j++) {
        if (a[j] > i)
          sum += INT_MIN;
        else
          sum += a[j];
        best = min(best, sum);
        ans = max(ans, (sum - best) - i);
      }
    }
    cout << ans << endl;
  }
}
