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;
char local = 'O';
template <typename T>
void read(T &a) {
  cin >> a;
}
template <typename T, typename T1>
void read(T &a, T1 &b) {
  cin >> a >> b;
}
template <typename T, typename T1, typename T2>
void read(T &a, T1 &b, T2 &c) {
  cin >> a >> b >> c;
}
template <typename T, typename T1>
void read(pair<T, T1> &p) {
  cin >> p.first >> p.second;
}
template <typename T>
void read(T arr[], long long n) {
  for (long long i = 0; i < n; i++) {
    read(arr[i]);
  }
}
template <typename T>
void read(vector<T> &arr) {
  for (long long i = 0; i < arr.size(); i++) {
    read(arr[i]);
  }
}
long long modularExponentiation(long long x, long long n, long long M) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return modularExponentiation((x * x) % M, n / 2, M);
  return (x * modularExponentiation((x * x) % M, (n - 1) / 2, M)) % M;
}
long long binaryExponentiation(long long x, long long n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return binaryExponentiation(x * x, n / 2);
  else
    return x * binaryExponentiation(x * x, (n - 1) / 2);
}
long long GCD(long long A, long long B) {
  if (B == 0)
    return A;
  else
    return GCD(B, A % B);
}
long long extended_GCD(long long a, long long b, long long &x, long long &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = extended_GCD(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return gcd;
}
long long modinv(long long a, long long mod) {
  long long x, y;
  extended_GCD(a, mod, x, y);
  if (x < 0) x += mod;
  return x;
}
long long modPrimeInverse(long long A, long long M) {
  return modularExponentiation(A, M - 2, M);
}
vector<long long> prime;
void sieve(long long N) {
  bool isPrime[N + !1];
  for (long long i = 0; i <= N; ++i) {
    isPrime[i] = true;
  }
  isPrime[0] = false;
  isPrime[1] = false;
  for (long long i = 2; i * i <= N; ++i) {
    if (isPrime[i] == true) {
      for (long long j = i * i; j <= N; j += i) isPrime[j] = false;
    }
  }
  for (long long i = 2; i <= N; i++) {
    if (isPrime[i] == true) {
      prime.push_back(i);
    }
  }
}
long long minPrime[1000007];
void factorSieve(long long n) {
  memset(minPrime, 0, sizeof(minPrime));
  for (long long i = 2; i * i <= n; ++i) {
    if (minPrime[i] == 0) {
      for (long long j = i * i; j <= n; j += i) {
        if (minPrime[j] == 0) {
          minPrime[j] = i;
        }
      }
    }
  }
  for (long long i = 2; i <= n; ++i) {
    if (minPrime[i] == 0) {
      minPrime[i] = i;
    }
  }
}
long long factorize(long long n) {
  map<long long, long long> m;
  m.clear();
  while (n != 1) {
    m[minPrime[n]]++;
    n /= minPrime[n];
  }
  for (auto i : m) {
    if (i.second > 1) {
      return 0;
    }
  }
  return 1;
}
void merge(long long A[], long long start, long long mid, long long end) {
  long long p = start, q = mid + 1;
  long long Arr[end - start + 1], k = 0;
  for (long long i = start; i <= end; i++) {
    if (p > mid)
      Arr[k++] = A[q++];
    else if (q > end)
      Arr[k++] = A[p++];
    else if (A[p] < A[q])
      Arr[k++] = A[p++];
    else
      Arr[k++] = A[q++];
  }
  for (long long p = 0; p < k; p++) {
    A[start++] = Arr[p];
  }
}
void merge_sort(long long A[], long long start, long long end) {
  if (start < end) {
    long long mid = (start + end) / 2;
    merge_sort(A, start, mid);
    merge_sort(A, mid + 1, end);
    merge(A, start, mid, end);
  }
}
long long binarySearch(long long a[], long long low, long long high,
                       long long key) {
  while (low <= high) {
    long long mid = (low + high) / 2;
    if (a[mid] < key) {
      low = mid + 1;
    } else if (a[mid] > key) {
      high = mid - 1;
    } else {
      return mid;
    }
  }
  return -1;
}
bool sortbysec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  if (a.first == b.first) {
    a.second > b.second;
  }
  return (a.first < b.first);
}
void fib(long long n, long long m, long long a[]) {
  if (n == 0) {
    a[0] = 0;
    a[1] = 1;
    return;
  }
  fib(n / 2, m, a);
  long long c, d;
  c = (a[0] * (2 * a[1] - a[0] + m)) % m;
  d = (a[0] * a[0] + a[1] * a[1]) % m;
  if (n % 2 == 0) {
    a[0] = c;
    a[1] = d;
  } else {
    a[0] = d;
    a[1] = c + d;
  }
}
struct comp {
  bool operator()(pair<long long, long long> a, pair<long long, long long> b) {
    if (a.first == b.first) {
      return a.second > b.second;
    }
    return a.first < b.first;
  }
};
double f(double x) { return log10(x + sqrt(10523)); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  long long a[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long maxi = 0;
  for (long long i = 0; i <= 30; i++) {
    long long r = 0;
    for (long long j = 0; j < n; j++) {
      if (a[j] <= i)
        r += a[j];
      else
        r = 0;
      r = max(r, 0LL);
      maxi = max(maxi, r - i);
    }
  }
  cout << maxi << "\n";
}
