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 no_of_elements;
  long long p, q, r;
  scanf("%d %I64d %I64d %I64d", &no_of_elements, &p, &q, &r);
  vector<int> A(no_of_elements + 1);
  for (int i = 1; i <= no_of_elements; i++) scanf("%d", &A[i]);
  vector<int> left_min(no_of_elements + 1);
  vector<int> left_max(no_of_elements + 1);
  left_min[1] = left_max[1] = A[1];
  for (int i = 2; i <= no_of_elements; i++) {
    left_min[i] = (left_min[i - 1] < A[i] ? left_min[i - 1] : A[i]);
    left_max[i] = (left_max[i - 1] > A[i] ? left_max[i - 1] : A[i]);
  }
  vector<int> right_min(no_of_elements + 1);
  vector<int> right_max(no_of_elements + 1);
  right_min[no_of_elements] = right_max[no_of_elements] = A[no_of_elements];
  for (int i = no_of_elements - 1; i >= 1; i--) {
    right_max[i] = (right_max[i + 1] > A[i] ? right_max[i + 1] : A[i]);
    right_min[i] = (right_min[i + 1] < A[i] ? right_min[i + 1] : A[i]);
  }
  long long ans = p * A[1] + q * A[1] + r * A[1];
  for (int i = 1; i <= no_of_elements; i++) {
    long long y = q * A[i];
    long long x = (p > 0 ? p * left_max[i] : p * left_min[i]);
    long long z = (r > 0 ? r * right_max[i] : r * right_min[i]);
    ans = (ans > x + y + z ? ans : x + y + z);
  }
  printf("%I64d\n", ans);
  return 0;
}
