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 maxn = 1e5 + 100;
long long dp1[maxn][2];
long long dp2[maxn][2];
long long n, p, q, r;
long long a[100000 + 100];
int main() {
  cin >> n >> p >> q >> r;
  for (long long y = 0; y < n; y++) cin >> a[y];
  dp1[0][0] = p * a[0];
  dp1[0][1] = dp1[0][0] + 1LL * q * a[0];
  dp2[0][0] = 1LL * p * a[0];
  dp2[0][1] = dp1[0][0] + 1LL * q * a[0];
  for (long long y = 1; y < n; y++) {
    dp1[y][0] = 1LL * a[y] * p;
    dp2[y][0] = max(dp2[y - 1][0], dp1[y][0]);
  }
  for (long long y = 1; y < n; y++) {
    dp1[y][1] = 1LL * a[y] * q + dp2[y][0];
    dp2[y][1] = max(dp2[y - 1][1], dp1[y][1]);
  }
  long long maxs = 1LL * dp2[0][1] + 1LL * r * a[0];
  for (long long y = 0; y < n; y++) {
    maxs = max(maxs, 1LL * dp2[y][1] + 1LL * r * a[y]);
  }
  cout << maxs;
}
