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() {
  long long int n, p, q, r, a[100000], s = -3000000000000000000, dpmi[100000],
                                       dpma[100000], dppmi[100000],
                                       dppma[100000], i;
  cin >> n >> p >> q >> r;
  for (i = 0; i < n; i++) cin >> a[i];
  dpmi[0] = a[0];
  dpma[0] = a[0];
  dppmi[n - 1] = a[n - 1];
  dppma[n - 1] = a[n - 1];
  for (i = 1; i < n; i++) {
    dpma[i] = max(dpma[i - 1], a[i]);
    dpmi[i] = min(dpmi[i - 1], a[i]);
  }
  for (i = n - 2; i >= 0; i--) {
    dppma[i] = max(dppma[i + 1], a[i]);
    dppmi[i] = min(dppmi[i + 1], a[i]);
  }
  for (i = 0; i < n; i++) {
    if (p > 0) {
      if (r > 0) {
        s = max(s, p * dpma[i] + q * a[i] + r * dppma[i]);
      } else {
        s = max(s, p * dpma[i] + q * a[i] + r * dppmi[i]);
      }
    } else {
      if (r > 0) {
        s = max(s, p * dpmi[i] + q * a[i] + r * dppma[i]);
      } else {
        s = max(s, p * dpmi[i] + q * a[i] + r * dppmi[i]);
      }
    }
  }
  cout << s;
}
