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.
#include <bits/stdc++.h>
using namespace std;
long long inf = (1e9) + 7;
long long n, p, r, q;
long long tabela[100005][3];
vector<long long> v;
long long dp(int index, int qual) {
  if (qual == 3) return 0;
  if (index == n) return -inf * inf * 3 - (100);
  if (tabela[index][qual] != -inf * inf * 4) return tabela[index][qual];
  long long aux = -inf * inf * 3 - 100;
  if (qual == 0) {
    aux = max(aux, dp(index + 1, qual));
    aux = max(aux, p * v[index] + dp(index, qual + 1));
  }
  if (qual == 1) {
    aux = max(aux, dp(index + 1, qual));
    aux = max(aux, q * v[index] + dp(index, qual + 1));
  }
  if (qual == 2) {
    aux = max(aux, dp(index + 1, qual));
    aux = max(aux, r * v[index] + dp(index, qual + 1));
  }
  tabela[index][qual] = aux;
  return aux;
}
int main() {
  for (int i = 0; i < 100005; ++i) {
    for (int j = 0; j < 3; ++j) {
      tabela[i][j] = -inf * inf * 4;
    }
  }
  cin >> n >> p >> q >> r;
  long long aux;
  for (int i = 0; i < n; ++i) {
    scanf("%lld", &aux);
    v.push_back(aux);
  }
  printf("%lld\n", dp(0, 0));
}
