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 n, d[200005], p, q, r, bs[200005], sn[200005];
long long int cev;
vector<long long int> v;
int main() {
  scanf("%lld %lld %lld %lld", &n, &p, &q, &r);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &d[i]);
  }
  cev = d[0] * (p + q + r);
  bs[0] = d[0] * p;
  for (int i = 1; i < n; i++)
    if (d[i] * p > bs[i - 1])
      bs[i] = d[i] * p;
    else
      bs[i] = bs[i - 1];
  sn[n - 1] = d[n - 1] * r;
  for (int i = n - 2; i >= 0; i--)
    if (d[i] * r > sn[i + 1])
      sn[i] = d[i] * r;
    else
      sn[i] = sn[i + 1];
  for (int i = 0; i < n; i++) cev = max(cev, bs[i] + sn[i] + d[i] * q);
  cout << cev << endl;
}
