Codeforces 855B：Marvolo Gaunt's Ring（枚举，前后缀）

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·ai + q·aj + r·ak for given p, q, r and array a1, a2, ... an 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, ... an ( - 109 ≤ ai ≤ 109).

Output

Output a single integer the maximum value of p·ai + q·aj + r·ak 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.

题意

给出一个有n个数的序列a[1],a[2]……a[n]，使得在i<=j<=k的条件下，令p*a[i]+q*a[j]+r*a[k]的值最大并输出这个值

思路

维护一个位置的前后缀left和right，left[i]表示位置i之前的元素与p相乘的最大值，right表示位置i之后的元素与q相乘的最大值，然后枚举每个位置，计算max(left[i]+q*a[i]+right([i])

代码

#include <bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define ms(a,b) memset(a,b,sizeof(a))
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int maxn=1e6+10;
const int mod=1e9+7;
const int maxm=1e3+10;
using namespace std;
ll Left[maxn];
ll Right[maxn];
ll a[maxn];
int main(int argc, char const *argv[])
{
    #ifndef ONLINE_JUDGE
        freopen("/home/wzy/in.txt", "r", stdin);
        freopen("/home/wzy/out.txt", "w", stdout);
        srand((unsigned int)time(NULL));
    #endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    ll p,q,r;
    cin>>n>>p>>q>>r;
    for(int i=0;i<n;i++)
        cin>>a[i];
    Left[0]=a[0]*p;
    Right[n-1]=a[n-1]*r;
    for(int i=1;i<n;i++)
        Left[i]=max(Left[i-1],p*a[i]);
    for(int i=n-2;i>=0;i--)
        Right[i]=max(Right[i+1],r*a[i]);
    ll ans=-INF;
    for(int i=0;i<n;i++)
        ans=max(ans,Left[i]+q*a[i]+Right[i]);
    cout<<ans<<endl;
    #ifndef ONLINE_JUDGE
        cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
    #endif
    return 0;
}