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 ar[100007], mx = -1e9, mn = 1e9;
struct st {
  long long int mx, mn;
} before[100007];
struct stt {
  long long int mx, mn;
} after[100007];
int main() {
  long long int n, p, q, r;
  cin >> n >> p >> q >> r;
  for (long long int i = 0; i < n; i++) {
    cin >> ar[i];
  }
  for (long long int i = 0; i < n; i++) {
    mn = min(ar[i], mn);
    mx = max(mx, ar[i]);
    before[i].mn = mn;
    before[i].mx = mx;
  }
  mn = 1e9;
  mx = ar[n - 1];
  for (long long int i = n - 1; i >= 0; i--) {
    mn = min(ar[i], mn);
    mx = max(mx, ar[i]);
    after[i].mn = mn;
    after[i].mx = mx;
  }
  long long int ans;
  for (long long int i = 0; i < n; i++) {
    long long int a = p * ar[i] + q * ar[i] + r * ar[i];
    long long int b = max(p * before[i].mn, p * before[i].mx) + (q * ar[i]) +
                      max(r * after[i].mn, r * after[i].mx);
    if (i == 0)
      ans = max(a, b);
    else
      ans = max(ans, max(a, b));
  }
  cout << ans << endl;
}
