CodeForces - 855B Marvolo Gaunt's Ring（DP）

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.

思路：我忽略了题目中的1 ≤ i ≤ j ≤ k ≤ n，导致一直wa，满脸泪水。

#include <cstdio>
#include <iostream>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
using namespace std;
#define ll long long

ll p, q, r, num[100000+8], miao1 = -8e18, miao2 = -8e18, miao3 = -8e18, ans;
int n;

int main()
{
    scanf("%d%lld%lld%lld", &n, &p, &q, &r);
    for(int i = 0; i<n; i++)
    {
        scanf("%lld", &num[i]);
        miao1 = max(miao1, p*num[i]);
        miao2 = max(miao2, miao1+q*num[i]);//保证1 ≤ i ≤ j 
        miao3 = max(miao3, miao2+r*num[i]);//保证1 ≤ i ≤ j ≤ k ≤ n
    }
    printf("%lld\n", miao3);
    return 0;
}