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 MAXN = 180;
const int MAXK = 80;
void solve() {
  int n;
  cin >> n;
  long long p, q, r;
  cin >> p >> q >> r;
  vector<long long> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  vector<long long> dp1(n);
  for (int i = 0; i < n; ++i) dp1[i] = a[i] * p;
  vector<long long> dp2(n);
  long long prevDpMax = -1e19;
  for (int j = 0; j < n; ++j) {
    prevDpMax = max(prevDpMax, dp1[j]);
    dp2[j] = prevDpMax + a[j] * q;
  }
  vector<long long> dp3(n);
  prevDpMax = -1e19;
  for (int k = 0; k < n; ++k) {
    prevDpMax = max(prevDpMax, dp2[k]);
    dp3[k] = prevDpMax + a[k] * r;
  }
  long long ans = -1e19;
  for (int i = 0; i < n; ++i) ans = max(ans, dp3[i]);
  cout << ans << endl;
}
int main() {
  cin.sync_with_stdio(false);
  cout.sync_with_stdio(false);
  solve();
  return 0;
}
