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：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <map>
 7 #define ll long long
 8 using namespace std;
 9 const ll inf = -4*1e18-5;
10
11 int main() {
12     int n;
13     scanf("%d",&n);
14     ll p,q,r;
15     scanf("%lld%lld%lld",&p,&q,&r);
16     ll u,m1,m2,m3;
17 //    printf("%lld ",inf);
18      m1 = m2 = m3 =inf;
19     while(n--) {
20         scanf("%lld",&u);
21         m1 = max(m1,u*p);
22         m2 = max(m2,u*q + m1);
23         m3 = max(m3,u*r + m2);
24     }
25     printf("%lld\n",m3);
26     return 0;
27 }