## 题目：

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$p， q， r$，给出n个数v[i]，问$\max(p∗ai+q∗aj+r∗ak)\ \ (i≤j≤k)$

## 思路：

枚举$a_j$，当$0≤p$，选取$max(ai),i∈[1,j]$, 否则选取$min(ai),i∈[1,j]$。

q同理。对于取max和取min，维护前缀和后缀即可。

## 实现：

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const ll maxn = int(1e5)+7, INF = 0x7f7f7f7f7f7f7f7f;
ll n, p, q, r;
ll premi[maxn], prema[maxn], sufmi[maxn], sufma[maxn], num[maxn];
int main() {
scanf("%lld%lld%lld%lld", &n, &p, &q, &r);
for(int i=1 ; i<=n ; i++) scanf("%lld", num+i);
prema[0] = sufma[n+1] = -INF;
premi[0] = sufmi[n+1] = INF;
for(int i=n ; i>=1 ; i--) {
sufma[i] = max(num[i], sufma[i+1]);
sufmi[i] = min(num[i], sufmi[i+1]);
}
for(int i=1 ; i<=n ; i++) {
prema[i] = max(num[i], prema[i-1]);
premi[i] = min(num[i], premi[i-1]);
}
ll ans = -INF, tmp;
for(int i=1 ; tmp = 0, i<=n ; i++) {
tmp += p * (p < 0 ? premi[i] : prema[i]);
tmp += r * (r < 0 ? sufmi[i] : sufma[i]);
ans = max(ans, tmp + q * num[i]);
}
printf("%lld\n", ans);
return 0;
}