B 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 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 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.

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <string>
#include <cstring>
#include <cstdlib>
#include <map>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
typedef long long lli;
using namespace std;
const int  mxn = 1e9;
#define TLE std::ios::sync_with_stdio(false);   cin.tie(NULL);   cout.tie(NULL);   cout.precision(10);
multiset<int> v;
multiset<int> :: iterator it ;
int main()
{
    TLE;
    lli n,cnt=1,m,k,ll;
    while(cin>>n>>m>>k>>ll)
    {
        int flag1 = 0,flag2=0;lli mx,x;
        if(m>=0 && k>=0 && ll>=0 )
            flag1=1;
        if(m<0 && k<0 && ll<0 )
            flag2=1;
        if(flag1)
        {
            cin>>mx;
            for(int i=2; i<=n; i++)
            {
                cin>>x;
                mx = max( mx ,x );
            }
            cout<<(m+k+ll)*mx<<endl;
        }
        else if(flag2)
        {
            cin>>mx;
            for(int i=2; i<=n; i++)
            {
                cin>>x;
                mx = min( mx ,x );
            }
            cout<<(m+k+ll)*mx<<endl;
        }
        else
        {

            lli dp[n+5];
            lli l[n+5],r[n+5];
            for(int i=1;i<=n;i++)
            {
                cin>>dp[i];
            }
            l[1] = m*dp[1];
            for(int i=2; i<=n; i++)
            {
                l[i] = max( l[i-1] ,m*dp[i] );
            }
            r[n] = ll*dp[n];
            for(int i=n-1; i>=1; i--)
            {
                r[i] = max( r[i+1] , ll*dp[i] );
            }
            mx = -1000000000000000000000-5;
            for(int i=1; i<=n; i++)
                mx = max( l[i]+r[i]+dp[i]*k , mx);
            cout<<mx<<endl;
        }
    }
    return 0;
}

B Marvolo Gaunt's Ring

原文地址：https://www.cnblogs.com/Shallow-dream/p/11623521.html