Professor Dumbledore is helping Harry destroy the Horcruxes. He went to Gaunt Shack as he suspected a Horcrux to be present there. He saw Marvolo Gaunt's Ring and identified it as a Horcrux. Although he destroyed it, he is still affected by its curse. Professor Snape is helping Dumbledore remove the curse. For this, he wants to give Dumbledore exactly x drops of the potion he made. 

Value of x is calculated as maximum of p·ai + q·aj + r·ak for given p, q, r and array a1, a2, ... an such that 1 ≤ i ≤ j ≤ k ≤ n. Help Snape find the value of x. Do note that the value of x may be negative.

Input

First line of input contains 4 integers n, p, q, r ( - 109 ≤ p, q, r ≤ 109, 1 ≤ n ≤ 105).

Next line of input contains n space separated integers a1, a2, ... an ( - 109 ≤ ai ≤ 109).

Output

Output a single integer the maximum value of p·ai + q·aj + r·ak that can be obtained provided 1 ≤ i ≤ j ≤ k ≤ n.

Examples

Input

5 1 2 3
1 2 3 4 5


Output

30


Input

5 1 2 -3
-1 -2 -3 -4 -5


Output

12

Note

In the first sample case, we can take i = j = k = 5, thus making the answer as 1·5 + 2·5 + 3·5 = 30.

In second sample case, selecting i = j = 1 and k = 5 gives the answer 12.
#include <bits/stdc++.h>
using namespace std;
long long n, p, q, r;
long long t[100001];
pair<long long, long long> jobbra[100001], balra[100001];
long long what_if(long long index) {
  long long res = 0;
  if (p < 0)
    res += p * balra[index].first;
  else
    res += p * balra[index].second;
  if (r < 0)
    res += r * jobbra[index].first;
  else
    res += r * jobbra[index].second;
  return q * t[index] + res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> p >> q >> r;
  long long lk = INT_MAX, ln = INT_MIN;
  for (int i = 1; i <= n; i++) {
    long long be;
    cin >> t[i];
  }
  balra[1] = {t[1], t[1]};
  jobbra[n] = {t[n], t[n]};
  for (int i = 2; i <= n; i++) {
    balra[i].first = min(balra[i - 1].first, t[i]);
    balra[i].second = max(balra[i - 1].second, t[i]);
  }
  for (int i = n - 1; i >= 1; i--) {
    jobbra[i].first = min(jobbra[i + 1].first, t[i]);
    jobbra[i].second = max(jobbra[i + 1].second, t[i]);
  }
  long long best = what_if(1);
  for (int i = 1; i <= n; i++) best = max(best, what_if(i));
  cout << best;
}
