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;
const long long INF = 4e18;
long long mx[2][100010], mi[2][100010];
long long a[100010];
int main() {
  int n;
  scanf("%d", &n);
  long long p, q, r;
  scanf("%I64d%I64d%I64d", &p, &q, &r);
  for (int i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  mi[0][0] = mi[1][n + 1] = INF;
  mx[0][0] = mx[1][n + 1] = -INF;
  for (int i = 1; i <= n; i++) {
    mx[0][i] = max(mx[0][i - 1], a[i]);
    mi[0][i] = min(mi[0][i - 1], a[i]);
  }
  for (int i = n; i >= 1; i--) {
    mx[1][i] = max(mx[1][i + 1], a[i]);
    mi[1][i] = min(mi[1][i + 1], a[i]);
  }
  long long ans = -INF;
  for (int i = 1; i <= n; i++) {
    long long tmp = q * a[i];
    if (p < 0)
      tmp += p * mi[0][i];
    else
      tmp += p * mx[0][i];
    if (r < 0)
      tmp += r * mi[1][i];
    else
      tmp += r * mx[1][i];
    ans = max(ans, tmp);
  }
  printf("%I64d\n", ans);
  return 0;
}
