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;
void debug_vec(vector<long long>& v) {
  cout << '[';
  for (auto i : v) cout << i << ", ";
  cout << "]";
  cout << "\n";
}
void debug_set(set<long long>& st) {
  cout << "{";
  for (auto i : st) cout << i << ", ";
  cout << "}";
  cout << "\n";
}
void fast() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
bool prime[1000001];
void sieve() {
  memset(prime, true, sizeof(prime));
  prime[1] = false;
  for (int p = 2; p * p <= 1000000; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= 1000000; i += p) prime[i] = false;
    }
  }
}
void print(bool f) {
  if (f)
    cout << "YES\n";
  else
    cout << "NO\n";
}
void google(int x) { cout << "Case #" << x << ": "; }
long long binary_expo(long long a, long long b) {
  long long res = 1;
  while (b != 0) {
    if (b & 1) {
      res = (res * a) % 1000000007;
    }
    b >>= 1;
    a = (a * a) % 1000000007;
  }
  return res;
}
void solve() {
  long long n;
  long double p, q, r;
  cin >> n >> p >> q >> r;
  long double pre[n];
  long double suf[n];
  long double a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    pre[i] = LDBL_MIN;
    suf[i] = LDBL_MIN;
  }
  long double res = -1e+19;
  for (int i = 0; i < n; i++) {
    pre[i] = p * a[i];
    if (i >= 1) {
      pre[i] = max(pre[i], pre[i - 1]);
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    suf[i] = r * a[i];
    if (i + 1 < n) suf[i] = max(suf[i], suf[i + 1]);
  }
  for (int i = 0; i < n; i++) {
    res = max(res, pre[i] + suf[i] + q * a[i]);
  }
  cout << fixed << setprecision(0) << res << '\n';
}
int main() {
  fast();
  solve();
  return 0;
}
