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 n;
long long p, q, r;
long long arr[200000];
long long premax[200000];
long long premin[200000];
long long sufmax[200000];
long long sufmin[200000];
int main() {
  if (fopen("test.in", "r")) {
    freopen("test.in", "r", stdin);
  }
  cin >> n >> p >> q >> r;
  for (int i = 0; i < n; i++) cin >> arr[i];
  premax[0] = arr[0];
  premin[0] = arr[0];
  sufmax[0] = arr[n - 1];
  sufmin[0] = arr[n - 1];
  for (int i = 1; i < n; i++)
    premax[i] = max(arr[i], premax[i - 1]),
    premin[i] = min(arr[i], premin[i - 1]);
  for (int i = 1; i < n; i++) sufmax[i] = max(arr[n - i - 1], sufmax[i - 1]);
  for (int i = 1; i < n; i++) sufmin[i] = min(arr[n - i - 1], sufmin[i - 1]);
  long long ans = p * arr[0] + q * arr[0] + r * arr[0];
  for (int i = 0; i < n; i++) {
    long long cand = q * arr[i];
    if (p > 0)
      cand += premax[i] * p;
    else
      cand += premin[i] * p;
    if (r > 0)
      cand += sufmax[n - i - 1] * r;
    else
      cand += sufmin[n - i - 1] * r;
    ans = max(cand, ans);
  }
  cout << ans << '\n';
  return 0;
}
