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 int arr1[100005];
long long int arr[100007][3];
int main() {
  long long int i, j, n, p, q, r;
  cin >> n >> p >> q >> r;
  for (i = 0; i < n; i++) {
    cin >> arr1[i];
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < 3; j++) {
      if (i == 0) {
        if (j == 0)
          arr[i][j] = arr1[i] * p;
        else {
          if (j == 1)
            arr[i][j] = (arr[i][j - 1] + arr1[i] * q);
          else
            arr[i][j] = ((arr[i][j - 1] + arr1[i] * r));
        }
      } else {
        if (j == 0) {
          arr[i][j] = max(arr[i - 1][j], (arr1[i] * p));
        }
        if (j == 1) {
          arr[i][j] = max(arr[i - 1][j], (arr[i][j - 1] + arr1[i] * q));
        }
        if (j == 2) {
          arr[i][j] = max(arr[i - 1][j], (arr[i][j - 1] + arr1[i] * r));
        }
      }
    }
  }
  cout << arr[n - 1][2] << "\n";
  return 0;
}
