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 main() {
  int n, p, q, r;
  cin >> n;
  cin >> p;
  cin >> q;
  cin >> r;
  long long int arr1[n];
  long long int prefixmax[n];
  long long int suffixmax[n];
  for (int i = 0; i < n; i++) {
    cin >> arr1[i];
  }
  if (p == 0) {
    for (int i = 0; i < n; i++) {
      prefixmax[i] = 0;
    }
  } else {
    prefixmax[0] = arr1[0];
    for (int i = 1; i < n; i++) {
      prefixmax[i] = max(p * prefixmax[i - 1], p * arr1[i]) / p;
    }
  }
  if (r == 0) {
    for (int i = 0; i < n; i++) {
      suffixmax[i] = 0;
    }
  } else {
    suffixmax[n - 1] = arr1[n - 1];
    for (int i = n - 2; i >= 0; i--) {
      suffixmax[i] = max(r * suffixmax[i + 1], r * arr1[i]) / r;
    }
  }
  long long int sum = (p * prefixmax[0]) + (q * arr1[0]) + (r * suffixmax[0]);
  for (int i = 0; i < n; i++) {
    sum = max(sum, ((p * prefixmax[i]) + (q * arr1[i]) + (r * suffixmax[i])));
  }
  cout << sum;
}
