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;
priority_queue<long long, vector<long long>, greater<long long>> pq1, pq2;
long long power(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b % 2 != 0) res = res * a;
    b /= 2;
    a = a * a;
  }
  return res;
}
long long pref[100010] = {0}, ansf[100005], ansf2[100005], ans[100005],
          seg[200005] = {0}, n;
stack<long long> st;
void build(long long *ar) {
  long long sum = 0;
  for (int i = n; i < 2 * n; ++i) {
    sum += ar[i - n + 1];
    pref[i - n + 1] = sum;
    seg[i] = sum;
  }
  for (int i = n - 1; i >= 1; --i) {
    seg[i] = max(seg[i << 1], seg[i << 1 | 1]);
  }
}
long long calc(long long a, long long b) {
  long long res = -1000000000000000000;
  for (a += n, b += n; a < b; a >>= 1, b >>= 1) {
    if (a % 2 == 1) {
      res = max(res, seg[a]);
      a++;
    }
    if (b % 2 == 1) {
      b--;
      res = max(res, seg[b]);
    }
  }
  return res;
}
long long solve(long long *ar, long long fl) {
  build(ar);
  for (int i = n; i >= 1; --i) {
    if (st.size() == 0) {
      st.push(i);
      ans[i] = -1;
    } else {
      while (st.size() > 0 && ar[i] >= ar[st.top()]) {
        st.pop();
      }
      if (st.size() == 0) {
        ans[i] = -1;
        st.push(i);
      } else {
        ans[i] = st.top();
        st.push(i);
      }
    }
  }
  if (fl == 0) {
    for (int i = n - 1; i >= 1; --i) {
      if (ans[i] == -1) {
        long long l = i + 1;
        long long h = n;
        long long c = calc(l - 1, h);
        long long st = 0;
        ansf[i] = max(st, c - pref[i]);
      } else {
        long long l = i + 1;
        long long h = ans[i] - 1;
        if (l > h) continue;
        long long c = calc(l - 1, h);
        long long st = 0;
        ansf[i] = max(st, c - pref[i]);
      }
    }
  } else {
    for (int i = n - 1; i >= 1; --i) {
      if (ans[i] == -1) {
        long long l = i + 1;
        long long h = n;
        long long c = calc(l - 1, h);
        long long st = 0;
        ansf2[i] = max(st, c - pref[i]);
      } else {
        long long l = i + 1;
        long long h = ans[i] - 1;
        if (l > h) continue;
        long long c = calc(l - 1, h);
        long long st = 0;
        ansf2[i] = max(st, c - pref[i]);
      }
    }
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long ar[100005], br[100005];
  cin >> n;
  for (int i = 1; i < n + 1; ++i) {
    cin >> ar[i];
  }
  solve(ar, 0);
  for (int i = 1; i < n + 3; ++i) {
    pref[i] = 0;
    ans[i] = 0;
  }
  while (st.size() > 0) st.pop();
  for (int i = 1; i < n + 1; ++i) {
    br[i] = ar[n - i + 1];
  }
  solve(br, 1);
  long long mx = 0;
  for (int i = 1; i < n + 1; ++i) {
    mx = max(mx, ansf[i] + ansf2[n - i + 1]);
  }
  cout << mx << endl;
  return 0;
}
