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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, p, q, r;
  cin >> n >> p >> q >> r;
  long long int arr[n];
  for (long long int i = 0; i < n; i++) cin >> arr[i];
  long long int dp1[n], dp2[n], dp3[n];
  dp1[0] = arr[0] * p;
  dp2[0] = dp1[0] + arr[0] * q;
  dp3[0] = dp2[0] + arr[0] * r;
  for (long long int i = 1; i < n; i++) {
    dp1[i] = max(dp1[i - 1], arr[i] * p);
    dp2[i] = max(dp2[i - 1], dp1[i] + arr[i] * q);
    dp3[i] = max(dp3[i - 1], dp2[i] + arr[i] * r);
  }
  cout << dp3[n - 1] << "\n";
  return 0;
}
