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;
int n;
vector<long long> arr;
long long elem[3];
long long dp[3][int(1e5 + 2)];
bool is_set[3][int(1e5 + 2)];
long long maximize(int stage, int k) {
  if (stage == 3 || k == n) return 0;
  if (is_set[stage][k]) return dp[stage][k];
  if (k < n) {
    long long t1 =
        arr[k] * elem[stage] + ((stage < 2) ? maximize(stage + 1, k) : 0);
    dp[stage][k] = t1;
    is_set[stage][k] = true;
  }
  if (k + 1 < n) {
    long long t2 = maximize(stage, k + 1);
    dp[stage][k] = (is_set[stage][k]) ? max(dp[stage][k], t2) : t2;
  }
  is_set[stage][k] = true;
  return dp[stage][k];
}
int main() {
  cin >> n;
  cin >> elem[0] >> elem[1] >> elem[2];
  arr.resize(n + 1);
  for (int i = 0; i < n; i++) cin >> arr[i];
  for (int i = 0; i < 3; i++)
    for (int j = 0; j < n; j++) is_set[i][j] = false;
  maximize(0, 0);
  cout << dp[0][0] << endl;
  return 0;
}
