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 int N = 100005;
int n;
long long p, q, r, a[N];
long long one[N], two[N], three[N];
int main() {
  scanf("%d %lld %lld %lld", &n, &p, &q, &r);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", a + i);
  }
  one[n + 1] = two[n + 1] = three[n + 1] = -8e18 - 10;
  for (int i = n; i; --i) {
    one[i] = r * a[i];
    one[i] = max(one[i], one[i + 1]);
  }
  for (int i = n; i; --i) {
    two[i] = q * a[i] + one[i];
    two[i] = max(two[i], two[i + 1]);
  }
  for (int i = n; i; --i) {
    three[i] = p * a[i] + two[i];
    three[i] = max(three[i], three[i + 1]);
  }
  printf("%lld\n", three[1]);
  return 0;
}
