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;
long long arr[100005];
long long tq[3];
long long dp[100005][3];
long long solve(long long i, long long id) {
  if (id == -1) return 0;
  if (dp[i][id] != LLONG_MIN) {
    return dp[i][id];
  }
  if (i == 0) {
    long long temp = 0;
    for (long long p = 0; p <= id; p++) temp += tq[p];
    return dp[i][id] = arr[i] * temp;
  }
  long long a = tq[id] * arr[i] + solve(i, id - 1);
  long long b = solve(i - 1, id);
  return dp[i][id] = max(a, b);
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tt = 1;
  while (tt--) {
    long long n, p, q, r;
    cin >> n >> p >> q >> r;
    tq[0] = p;
    tq[1] = q;
    tq[2] = r;
    for (long long i = 0; i < 100005; i++) {
      for (long long j = 0; j < 3; j++) dp[i][j] = LLONG_MIN;
    }
    for (long long i = 0; i < n; i++) {
      cin >> arr[i];
    }
    long long ans = solve(n - 1, 2);
    cout << ans;
    cout << '\n';
  }
  return 0;
}
