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 long long N1 = 1000010;
const long long N2 = 2000010;
const long long mod1 = 1e2 + 7;
const long long mod2 = 1e9 + 7;
long long a1[N1], p, q, r;
long long n;
long long leftMin[N1];
long long rightMin[N1];
long long leftMax[N1];
long long rightMax[N1];
long long maxProduct() {
  long long max_product = -3e18;
  leftMax[0] = a1[0];
  leftMin[0] = a1[0];
  for (int i = 1; i < n; i++) {
    leftMax[i] = max(leftMax[i - 1], a1[i]);
    leftMin[i] = min(leftMin[i - 1], a1[i]);
  }
  rightMax[n - 1] = a1[n - 1];
  rightMin[n - 1] = a1[n - 1];
  for (int j = n - 2; j >= 0; j--) {
    rightMax[j] = max(rightMax[j + 1], a1[j]);
    rightMin[j] = min(rightMin[j + 1], a1[j]);
  }
  for (int i = 0; i < n; i++) {
    long long max1 = max(q * a1[i] + p * leftMax[i] + r * rightMax[i],
                         q * a1[i] + p * leftMin[i] + r * rightMin[i]);
    long long max2 = max(q * a1[i] + p * leftMax[i] + r * rightMin[i],
                         q * a1[i] + p * leftMin[i] + r * rightMax[i]);
    max_product = max(max_product, max(max1, max2));
  }
  return max_product;
}
int main() {
  cin >> n >> p >> q >> r;
  ;
  for (int i = 0; i < n; i++) cin >> a1[i];
  ;
  long long x = maxProduct();
  cout << x << "\n";
  ;
  return 0;
}
