Marvolo Gaunt's Ring CodeForces

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 a₁, a₂, ... 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 a₁, a₂, ... 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.

Example

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.

题解：注意顺序是一定的，i<=j<=k，所以可以枚举中间值j，然后用两个数组分别存第一项的前缀最大值，第三项的后缀最大值，最后加起来遍历一遍就行了。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn=1e5+5;

ll n,p,q,r;
ll a[maxn],b[maxn],c[maxn];

int main()
{   cin>>n>>p>>q>>r;
    ll temp;
    for(int i=1;i<=n;i++){
        cin>>temp;
        a[i]=p*temp;
        b[i]=q*temp;
        c[i]=r*temp;
    }    
    temp=a[1];
    for(int i=1;i<=n;i++){
        temp=max(temp,a[i]);
        a[i]=temp;
    }
    temp=c[n];
    for(int i=n;i>=1;i--){
        temp=max(temp,c[i]);
        c[i]=temp;
    }
    temp=a[1]+b[1]+c[1];
    for(int i=1;i<=n;i++) temp=max(temp,a[i]+b[i]+c[i]);
    cout<<temp<<endl;
}