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.
import sys
import bisect
import heapq
import math
from functools import lru_cache
from collections import defaultdict,Counter


#sys.setrecursionlimit(10**6)
input = sys.stdin.readline
I = lambda: int(input())
S = lambda: input()
L = lambda: list(map(int, input().split()))
###################################################




def solution(n,p,q,r):
    a = L()
    temp1 = float('inf')
    temp2 = float('-inf')
    left_min = []
    left_max = []
    right_min = []
    right_max = []
    for i in a:
        temp1 = min(i,temp1)
        temp2 = max(i,temp2)
        left_min.append(temp1)
        left_max.append(temp2)
    temp1 = float('inf')
    temp2 = float('-inf')
    for i in a[::-1]:
        temp1 = min(i,temp1)
        temp2 = max(i,temp2)
        right_min.append(temp1)
        right_max.append(temp2)
    right_min.reverse()
    right_max.reverse()
    res = float('-inf')
    n = len(a)
    for i in range(n):
        x = max(p*left_max[i],p*left_min[i])
        y = max(r*right_max[i],r*right_min[i])
        z = q*a[i]
        res = max(res,x+y+z)
    print(res)
if __name__=='__main__':
    n,p,q,r = L()
    solution(n,p,q,r)