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 int mod = 1e9 + 7;
const int N = 1e5;
int tree[4 * N], tree1[4 * N];
int cnt;
int a[100050];
stack<int> stk, stk1;
int f[100050], g[100050], s[100050];
void pushup(int i) {
  tree[i] = max(tree[i << 1], tree[i << 1 | 1]);
  tree1[i] = min(tree1[i << 1], tree1[i << 1 | 1]);
}
void build(int i, int l, int r) {
  if (l == r) {
    tree1[i] = tree[i] = s[cnt++];
    return;
  }
  int mid = (l + r) / 2;
  build(i << 1, l, mid);
  build(i << 1 | 1, mid + 1, r);
  pushup(i);
}
int query(int i, int l, int r, int first, int second) {
  if (first <= l && r <= second) return tree[i];
  int maxm = -1e9;
  int mid = (l + r) / 2;
  if (first <= mid) maxm = max(maxm, query(i << 1, l, mid, first, second));
  if (second > mid)
    maxm = max(maxm, query(i << 1 | 1, mid + 1, r, first, second));
  return maxm;
}
int query1(int i, int l, int r, int first, int second) {
  if (first <= l && r <= second) return tree1[i];
  int maxm = 1e9;
  int mid = (l + r) / 2;
  if (first <= mid) maxm = min(maxm, query1(i << 1, l, mid, first, second));
  if (second > mid)
    maxm = min(maxm, query1(i << 1 | 1, mid + 1, r, first, second));
  return maxm;
}
int main() {
  int n;
  cin >> n;
  for (int i(0); i < n; i++) cin >> a[i];
  int L[100050], R[100050];
  for (int i = 0; i < n; i++) {
    while (!stk.empty() && a[stk.top()] <= a[i]) {
      stk.pop();
    }
    if (stk.empty()) {
      L[i] = -1;
    } else
      L[i] = stk.top();
    stk.push(i);
  }
  for (int i = n - 1; i >= 0; i--) {
    while (!stk1.empty() && a[stk1.top()] <= a[i]) {
      stk1.pop();
    }
    if (stk1.empty()) {
      R[i] = n;
    } else
      R[i] = stk1.top();
    stk1.push(i);
  }
  s[0] = a[0];
  for (int i(1); i < n; i++) s[i] = s[i - 1] + a[i];
  build(1, 1, n);
  memset(f, -0x3f, sizeof f);
  memset(g, -0x3f, sizeof g);
  for (int i(0); i < n; i++) {
    if (a[i] > 0) {
      if (i == 0)
        f[i] = 0;
      else {
        int p = query1(1, 1, n, L[i] + 2, i);
        f[i] = max(f[i], s[i - 1] - p);
        if (L[i] == -1) f[i] = max(f[i], s[i - 1]);
        f[i] = max(f[i], 0);
      }
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    if (a[i] > 0) {
      int p = query(1, 1, n, i + 2, R[i]);
      g[i] = max(g[i], p - s[i]);
      g[i] = max(g[i], 0);
    }
  }
  int ans = 0;
  for (int i(0); i < n; i++) {
    if (a[i] > 0) ans = max(ans, f[i] + g[i]);
  }
  cout << ans;
}
