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 mod = 1e9 + 7;
const long long int mod_ = 998244353;
const int INF = 1000000000;
const long long int INF64 = (long long int)1E18;
long long int n, p, q, r;
long long int dp0[100005], dp1[100005], dp2[100005];
vector<long long int> arr;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> p >> q >> r;
  arr.resize(n);
  long long int mini = INF64, maxi = -INF64;
  for (long long int& ele : arr) cin >> ele;
  dp0[0] = arr[0] * p;
  dp1[0] = dp0[0] + arr[0] * q;
  dp2[0] = dp1[0] + arr[0] * r;
  for (int i = 1; i < n; i++) dp0[i] = max(dp0[i - 1], arr[i] * p);
  for (int j = 1; j < n; j++) dp1[j] = max(dp1[j - 1], arr[j] * q + dp0[j]);
  for (int k = 1; k < n; k++) dp2[k] = max(dp2[k - 1], arr[k] * r + dp1[k]);
  cout << dp2[n - 1];
  return 0;
}
