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;
string vow = "aeiou";
const int dxhorse[] = {-2, -2, -1, -1, 1, 1, 2, 2};
const int dyhorse[] = {1, -1, 2, -2, 2, -2, 1, -1};
const int dx[] = {-1, 0, 0, 1};
const int dy[] = {0, -1, 1, 0};
const long double pie = 3.14159265358979;
const long long mod = 1e9 + 7;
void solve(int test_case) {
  int n;
  cin >> n;
  vector<long long> p(4);
  for (int i = 1; i < 4; i++) cin >> p[i];
  vector<long long> v(n + 1);
  for (int i = 1; i < n + 1; i++) cin >> v[i];
  vector<vector<long long> > dp(4, vector<long long>(n + 1));
  for (int i = 0; i < 4; i++) dp[i][0] = LLONG_MIN;
  for (int j = 1; j < n + 1; j++) {
    for (int i = 1; i < 4; i++)
      dp[i][j] = max(dp[i][j - 1], dp[i - 1][j] + p[i] * v[j]);
  }
  cout << dp[3][n];
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  for (int i = 0; i < t; i++) solve(i);
  return 0;
}
