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;
static const int INF = 0x3f3f3f3f;
static const long long INFL = 0x3f3f3f3f3f3f3f3fLL;
template <typename T, typename U>
static void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static void amax(T &x, U y) {
  if (x < y) x = y;
}
int main() {
  int n;
  int p;
  int q;
  int r;
  while (~scanf("%d%d%d%d", &n, &p, &q, &r)) {
    vector<int> a(n);
    for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
    vector<long long> maxR(n + 1);
    maxR[n] = -INFL;
    for (int i = n - 1; i >= 0; --i)
      maxR[i] = max(maxR[i + 1], (long long)r * a[i]);
    long long maxL = -INFL;
    long long ans = -INFL;
    for (int(i) = 0; (i) < (int)(n); ++(i)) {
      amax(maxL, (long long)p * a[i]);
      amax(ans, maxL + (long long)q * a[i] + maxR[i]);
    }
    printf("%lld\n", ans);
  }
  return 0;
}
