codeforce 855B

B. Marvolo Gaunt's Ring

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

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.

 

这题题意是给定p，q，r，要求从数组中选择 i，j，k  （i<=j<=k) 

使得 p*i + q*j + r*k 最大，就是道暴力模拟水题，我竟然没有想到T T，直接掉了80分。。。

哇，我真的是个智障。

附ac代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <map>
#define ll long long 
using namespace std;
const ll inf = -4*1e18-5;

int main() {
    int n;
    scanf("%d",&n);
    ll p,q,r;
    scanf("%lld%lld%lld",&p,&q,&r);
    ll u,m1,m2,m3;
//    printf("%lld ",inf);
     m1 = m2 = m3 =inf;
    while(n--) {
        scanf("%lld",&u);
        m1 = max(m1,u*p);
        m2 = max(m2,u*q + m1);
        m3 = max(m3,u*r + m2);
    }
    printf("%lld\n",m3);
    return 0;
}