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>
const int N = 100007;
using namespace std;
long long a[N];
long long memo[N][4];
long long p, q, r;
int n;
long long solve(int i, int cnt) {
  if (i == n) {
    if (cnt == 0)
      return 0;
    else
      return -4000000000000000000;
  }
  if (memo[i][cnt] == 4000000000000000000) {
    if (cnt == 3) {
      memo[i][cnt] = max(solve(i + 1, cnt), p * a[i] + solve(i, cnt - 1));
    } else if (cnt == 2) {
      memo[i][cnt] = max(solve(i + 1, cnt), q * a[i] + solve(i, cnt - 1));
    } else if (cnt == 1) {
      memo[i][cnt] = max(solve(i + 1, cnt), r * a[i] + solve(i, cnt - 1));
    } else if (cnt == 0)
      memo[i][cnt] = solve(i + 1, cnt);
  }
  return memo[i][cnt];
}
int main() {
  for (int i = 0; i < N; ++i) {
    memo[i][0] = memo[i][1] = memo[i][2] = memo[i][3] = 4000000000000000000;
  }
  cin >> n >> p >> q >> r;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  cout << solve(0, 3) << endl;
}
