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 binomialCoeff(long long int n, long long int k) {
  long long int C[n + 1][k + 1];
  long long int i, j;
  for (i = 0; i <= n; i++) {
    for (j = 0; j <= min(i, k); j++) {
      if (j == 0 || j == i)
        C[i][j] = 1;
      else
        C[i][j] = (C[i - 1][j - 1] % 1000000007 + C[i - 1][j] % 1000000007) %
                  1000000007;
    }
    C[i][j] = C[i][j] % 1000000007;
  }
  return C[n][k] % 1000000007;
}
void chatur() {
  long long int n, m, k = 0, p, q, r, s = 0, t = 0, l = 0, f = 0, g = 0,
                      max1 = 0, i;
  cin >> n;
  cin >> p >> q >> r;
  long long int a[n], b[n], c[n], d[n], bb[n][3];
  for (i = 0; i < n; i++) cin >> a[i];
  for (i = 0; i < n; i++) {
    b[i] = p * a[i];
    c[i] = q * a[i];
    d[i] = r * a[i];
  }
  bb[0][0] = b[0];
  bb[0][1] = c[0];
  bb[0][2] = d[0];
  for (i = 1; i < n; i++) {
    bb[i][0] = max(bb[i - 1][0], b[i]);
    bb[i][1] = max(bb[i - 1][1], c[i]);
    bb[i][2] = max(bb[i - 1][2], d[i]);
  }
  long long int pq[n];
  pq[0] = bb[0][0] + c[0];
  for (i = 1; i < n; i++) {
    pq[i] = max(pq[i - 1], c[i] + bb[i][0]);
  }
  max1 = d[0] + pq[0];
  for (i = 1; i < n; i++) {
    max1 = max(max1, d[i] + pq[i]);
  }
  cout << max1;
}
int main() {
  int t = 1;
  while (t--) {
    chatur();
  }
}
