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 int N = 100100;
long long n, b[4], a[N], minL[N], maxL[N], minR[N], maxR[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= 3; i++) cin >> b[i];
  minL[0] = 1e15;
  maxL[0] = -1e15;
  minR[n + 1] = 1e15;
  maxR[n + 1] = -1e15;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    minL[i] = min(minL[i - 1], a[i]);
    maxL[i] = max(maxL[i - 1], a[i]);
  }
  for (int i = n; i >= 1; i--) {
    minR[i] = min(minR[i + 1], a[i]);
    maxR[i] = max(maxR[i + 1], a[i]);
  }
  long long res = -4e18;
  for (int i = 1; i <= n; i++) {
    long long sum = b[2] * a[i];
    if (b[1] < 0)
      sum += minL[i] * b[1];
    else
      sum += maxL[i] * b[1];
    if (b[3] < 0)
      sum += minR[i] * b[3];
    else
      sum += maxR[i] * b[3];
    res = max(res, sum);
  }
  cout << res << endl;
  return 0;
}
