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, i, j, k, ans = 0, maxx, minn, ans1;
  cin >> n >> p >> q >> r;
  long long int A[n], B[n], C[n], D[n], E[n];
  for (i = 0; i < n; i++) cin >> A[i];
  maxx = minn = A[0];
  B[0] = A[0];
  C[0] = A[0];
  for (i = 1; i < n; i++) {
    if (maxx < A[i]) maxx = A[i];
    if (minn > A[i]) minn = A[i];
    B[i] = maxx;
    C[i] = minn;
  }
  maxx = minn = A[n - 1];
  D[n - 1] = A[n - 1];
  E[n - 1] = A[n - 1];
  for (i = n - 2; i >= 0; i--) {
    if (maxx < A[i]) maxx = A[i];
    if (minn > A[i]) minn = A[i];
    D[i] = maxx;
    E[i] = minn;
  }
  if (p >= 0 && r >= 0)
    ans = B[0] * p + A[0] * q + D[0] * r;
  else if (p >= 0 && r < 0)
    ans = B[0] * p + A[0] * q + E[0] * r;
  else if (p < 0 && r >= 0)
    ans = C[0] * p + A[0] * q + D[0] * r;
  else if (p < 0 && r < 0)
    ans = C[0] * p + A[0] * q + E[0] * r;
  for (i = 1; i < n; i++) {
    if (p >= 0 && r >= 0)
      ans1 = B[i] * p + A[i] * q + D[i] * r;
    else if (p >= 0 && r < 0)
      ans1 = B[i] * p + A[i] * q + E[i] * r;
    else if (p < 0 && r >= 0)
      ans1 = C[i] * p + A[i] * q + D[i] * r;
    else if (p < 0 && r < 0)
      ans1 = C[i] * p + A[i] * q + E[i] * r;
    if (ans < ans1) ans = ans1;
  }
  cout << ans << endl;
  return (0);
}
