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 long long maxn = 2e5 + 5;
long long n, m, x, y, z;
long long a[maxn], f[maxn], ff[maxn], g[maxn], gg[maxn];
const long long inf = 4e18;
int main() {
  cin >> n >> x >> y >> z;
  for (register long long i = 1; i <= n; ++i) cin >> a[i];
  for (register long long i = 0; i <= n + 1; ++i)
    gg[i] = g[i] = inf, ff[i] = f[i] = -inf;
  for (register long long i = 1; i <= n; ++i)
    f[i] = max(f[i - 1], a[i]), g[i] = min(g[i - 1], a[i]);
  for (register long long i = n; i >= 1; --i)
    ff[i] = max(ff[i + 1], a[i]), gg[i] = min(gg[i + 1], a[i]);
  long long ans = -inf;
  for (register long long i = 1; i <= n; ++i) {
    long long tmp = a[i] * y;
    if (x > 0)
      tmp += f[i] * x;
    else
      tmp += g[i] * x;
    if (z > 0)
      tmp += ff[i] * z;
    else
      tmp += gg[i] * z;
    ans = max(ans, tmp);
  }
  cout << ans << endl;
  return 0;
}
