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 MOD = pow(10, 9) + 7;
long long INF = 1ll << 60;
int main() {
  long long n, p, q, r;
  cin >> n >> p >> q >> r;
  long long a[n + 1];
  for (long long i = 1; i < n + 1; i++) cin >> a[i];
  long long dp[n + 1];
  for (long long i = 0; i < n + 1; i++) dp[i] = -9000000000000000000;
  for (long long i = 1; i < n + 1; i++) dp[i] = max(dp[i - 1], p * a[i]);
  for (long long j = 1; j < n + 1; j++)
    dp[j] = max(dp[j - 1], dp[j] + q * a[j]);
  for (long long k = 1; k < n + 1; k++)
    dp[k] = max(dp[k - 1], dp[k] + r * a[k]);
  cout << dp[n] << endl;
}
