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;
int max(int a, int b, int c) {
  long long testa = 1LL * a * c;
  long long testb = 1LL * b * c;
  if (testa > testb) return a;
  return b;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int p, q, r, n;
  cin >> n >> p >> q >> r;
  std::vector<int> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  if (n == 1) {
    cout << 1LL * (0LL + p + q + r) * v[0] << '\n';
    return 0;
  }
  std::vector<int> MaxR(n);
  MaxR.back() = v.back();
  for (int i = n - 2; i >= 0; i--) {
    MaxR[i] = max(MaxR[i + 1], v[i], r);
  }
  int MaxI = v[0];
  long long sol = -5e18;
  for (int i = 0; i < n; i++) {
    MaxI = max(MaxI, v[i], p);
    sol = max(sol, 1LL * p * MaxI + 1LL * v[i] * q + 1LL * MaxR[i] * r);
  }
  cout << sol << '\n';
  return 0;
}
