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 = 100007;
int A[N], ML[N], ml[N], MR[N], mr[N];
int main() {
  int n, p, q, r;
  cin >> n >> p >> q >> r;
  for (int i = 0; i < n; ++i) cin >> A[i];
  ML[0] = ml[0] = A[0];
  MR[n - 1] = mr[n - 1] = A[n - 1];
  for (int i = 1; i < n; ++i) {
    ml[i] = min(ml[i - 1], A[i]);
    ML[i] = max(ML[i - 1], A[i]);
  }
  for (int i = n - 2; i >= 0; --i) {
    mr[i] = min(mr[i + 1], A[i]);
    MR[i] = max(MR[i + 1], A[i]);
  }
  long long sum = -3e18, left, right;
  for (int i = 0; i < n; ++i) {
    left = 1ll * p * ((p < 0) ? ml[i] : ML[i]);
    right = 1ll * r * ((r < 0) ? mr[i] : MR[i]);
    sum = max(left + right + 1ll * q * A[i], sum);
  }
  cout << sum;
}
