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 n, p, q, r, i, a[int(1e5 + 1)], minl[int(1e5 + 1)],
      maxl[int(1e5 + 1)], minr[int(1e5 + 1)], maxr[int(1e5 + 1)];
  scanf("%lld%lld%lld%lld", &n, &p, &q, &r);
  for (i = 1; i <= n; i++) scanf("%lld", &a[i]);
  minl[1] = a[1];
  maxl[1] = a[1];
  for (i = 2; i <= n; i++) {
    minl[i] = minl[i - 1];
    maxl[i] = maxl[i - 1];
    if (a[i] < minl[i]) minl[i] = a[i];
    if (a[i] > maxl[i]) maxl[i] = a[i];
  }
  minr[n] = a[n];
  maxr[n] = a[n];
  for (i = n - 1; i >= 1; i--) {
    minr[i] = minr[i + 1];
    maxr[i] = maxr[i + 1];
    if (a[i] < minr[i]) minr[i] = a[i];
    if (a[i] > maxr[i]) maxr[i] = a[i];
  }
  if (p >= 0 && q >= 0 && r >= 0) {
    printf("%lld", (p + q + r) * maxl[n]);
    return 0;
  }
  if (p <= 0 && q <= 0 && r <= 0) {
    printf("%lld", (p + q + r) * minl[n]);
    return 0;
  }
  long long rez, t;
  for (i = 1; i <= n; i++) {
    t = q * a[i];
    if (p > 0)
      t += p * maxl[i];
    else
      t += p * minl[i];
    if (r > 0)
      t += r * maxr[i];
    else
      t += r * minr[i];
    if (t > rez || i == 1) rez = t;
  }
  printf("%lld", rez);
  return 0;
}
