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;
const int N = 1e5 + 1;
long long n, a[N], b[3], dpl[N], dpr[N], ans = -1e19 + 1;
int main() {
  cin >> n >> b[0] >> b[1] >> b[2];
  for (int i = 0; i < n; i++) cin >> a[i];
  dpl[0] = a[0] * b[0];
  for (int i = 1; i < n; i++) dpl[i] = max(dpl[i - 1], a[i] * b[0]);
  dpr[n - 1] = a[n - 1] * b[2];
  for (int i = n - 2; i >= 0; i--) dpr[i] = max(dpr[i + 1], a[i] * b[2]);
  for (int i = 0; i < n; i++) ans = max(ans, (dpr[i] + dpl[i] + (a[i] * b[1])));
  cout << ans;
}
