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 maxn = (1e5) + 1;
const long long int mod = 1000000007;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, x, y, z;
  cin >> n >> x >> y >> z;
  long long int dp[n][3];
  long long int i, j;
  long long int a[n];
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < 3; j++) {
      if (j == 0) {
        if (i == 0) {
          dp[i][j] = a[i] * x;
        } else {
          dp[i][j] = max(dp[i - 1][j], a[i] * x);
        }
      } else if (j == 1) {
        if (i == 0) {
          dp[i][j] = dp[i][0] + a[i] * y;
        } else {
          dp[i][j] = max(dp[i - 1][j], dp[i][0] + a[i] * y);
        }
      } else {
        if (i == 0) {
          dp[i][j] = dp[i][1] + a[i] * z;
        } else {
          dp[i][j] = max(dp[i - 1][j], dp[i][1] + a[i] * z);
        }
      }
    }
  }
  cout << dp[n - 1][2] << "\n";
  cerr << "\nTime elapsed:" << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}
