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 M = 1000000007;
const int MM = 998244353;
const long double PI = acos(-1);
long long power(long long b, long long e, long long m) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b % m, e / 2, m) % m;
  return power(b * b % m, e / 2, m);
}
long long power(long long b, long long e) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b, e / 2);
  return power(b * b, e / 2);
}
template <typename T, typename U>
static inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T &x, U y) {
  if (x < y) x = y;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
  return os << '(' << p.first << "," << p.second << ')';
}
int _runtimeTerror_() {
  long long n, p, q, r;
  cin >> n >> p >> q >> r;
  long long ans = -8000000000000000000;
  long long a[n];
  for (int i = 0; i < n; ++i) cin >> a[i];
  long long px[n], sx[n];
  px[0] = a[0] * p;
  sx[n - 1] = a[n - 1] * r;
  for (int i = 1; i < n; ++i) px[i] = max(px[i - 1], a[i] * p);
  for (int i = n - 2; i >= 0; --i) sx[i] = max(sx[i + 1], a[i] * r);
  for (int i = 0; i < n; ++i) amax(ans, a[i] * q + px[i] + sx[i]);
  cout << ans;
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) _runtimeTerror_();
  return 0;
}
