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;
struct A {
  long long right;
  long long left;
  long long val;
};
void update(vector<A> *v) {
  long long M = -5100000000000000000;
  (*v)[0].left = (*v)[0].val;
  (*v)[(*v).size() - 1].right = (*v)[(*v).size() - 1].val;
  for (int i = 1; i < (*v).size(); ++i) {
    (*v)[i].left = (((*v)[i - 1].left) > ((*v)[i].val) ? ((*v)[i - 1].left)
                                                       : ((*v)[i].val));
  }
  for (int i = (*v).size() - 2; i >= 0; --i) {
    (*v)[i].right = (((*v)[i + 1].right) > ((*v)[i].val) ? ((*v)[i + 1].right)
                                                         : ((*v)[i].val));
  }
}
int main() {
  int n;
  long long p, q, r;
  cin >> n >> p >> q >> r;
  int i1 = -1, i2 = -1, i3 = -1;
  vector<long long> v;
  vector<A> v1, v2, v3;
  v.resize(n);
  v1.resize(n);
  v2.resize(n);
  v3.resize(n);
  long long max1 = -5100000000000000000, max2 = -5100000000000000000,
            max3 = -5100000000000000000;
  for (int i = 0; i < n; ++i) {
    cin >> v[i];
    v1[i].val = p * v[i];
    v2[i].val = q * v[i];
    v3[i].val = r * v[i];
  }
  update(&v1);
  update(&v2);
  update(&v3);
  for (int i = 0; i < v2.size(); ++i) {
    if (v1[i].left + v2[i].val + v3[i].right > max1) {
      max1 = v1[i].left + v2[i].val + v3[i].right;
    }
  }
  cout << max1 << endl;
  return 0;
  cout << max1 + max2 + max3;
  return 0;
}
