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;
const long long int mx = 1e5 + 1, Mod = 1e9 + 7, inf = 15e17;
long long int n, b[4];
long long int dp[4][mx], a[mx];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> b[1] >> b[2] >> b[3];
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int w = 1; w < 4; w++) {
    dp[w][1] = 0;
    for (int i = 1; i <= w; i++) dp[w][1] += b[i] * a[1];
    for (int i = 2; i <= n; i++) {
      dp[w][i] = max(dp[w][i - 1], dp[w - 1][i - 1] + a[i] * b[w]);
      if (w > 1)
        dp[w][i] =
            max(dp[w][i], dp[w - 2][i - 1] + a[i] * b[w] + a[i] * b[w - 1]);
      if (w > 2)
        dp[w][i] = max(dp[w][i], dp[w - 3][i - 1] + a[i] * b[w] +
                                     a[i] * b[w - 1] + a[i] * b[w - 2]);
    }
  }
  cout << dp[3][n] << '\n';
  return 0;
}
