# Marvolo Gaunt’s Ring

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·a i + q·a j + r·a k for given p, q, r and array a 1, a 2, … a n 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 a 1, a 2, … a n ( - 109 ≤ a i ≤ 109).

Output

Output a single integer the maximum value of p·a i + q·a j + r·a k that can be obtained provided 1 ≤ i ≤ j ≤ k ≤ n.

Examples

input

5 1 2 31 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.

Solution:

#include <bits/stdc++.h>

using namespace std;

const int N = 1234567;

int a[N];
long long L[N], R[N];

int main() {
long long p, q, r;
int n;
cin >> n >> p >> q >> r;
for (int i = 0; i < n; i++) {
scanf("%d", a + i);
}
L[0] = p * a[0];
for (int i = 1; i < n; i++) {
L[i] = max(L[i - 1], p * a[i]);
}
R[n - 1] = r * a[n - 1];
for (int i = n - 2; i >= 0; i--) {
R[i] = max(R[i + 1], r * a[i]);
}
long long ans = (long long) (-9e18);
for (int i = 0; i < n; i++) {
ans = max(ans, L[i] + q * a[i] + R[i]);
}
cout << ans << endl;
return 0;
}