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 p, q, r, n, a[100100];
long long maxl[100100];
long long maxr[100100];
long long minl[100100];
long long minr[100100];
long long ans = -9223372036854775000ll, tmp1, tmp2, tmp3, tmp4;
long long ma(long long x, long long y) { return x > y ? x : y; }
long long mi(long long x, long long y) { return x < y ? x : y; }
void init() {
  scanf("%I64d %I64d %I64d %I64d", &n, &p, &q, &r);
  int i;
  for (i = 0; i <= n + 1; i++) {
    maxl[i] = maxr[i] = -2147483647;
    minl[i] = minr[i] = 2147483647;
  }
  for (i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    maxl[i] = ma(a[i], maxl[i - 1]);
    minl[i] = mi(a[i], minl[i - 1]);
  }
  for (i = n; i >= 1; i--) {
    maxr[i] = ma(a[i], maxr[i + 1]);
    minr[i] = mi(a[i], minr[i + 1]);
  }
}
int main() {
  init();
  int i;
  for (i = 1; i <= n; i++) {
    tmp1 = p * maxl[i] + q * a[i] + r * maxr[i];
    tmp2 = p * minl[i] + q * a[i] + r * maxr[i];
    tmp3 = p * maxl[i] + q * a[i] + r * minr[i];
    tmp4 = p * minl[i] + q * a[i] + r * minr[i];
    ans = ma(ans, ma(ma(tmp1, tmp2), ma(tmp3, tmp4)));
  }
  cout << ans;
  return 0;
}
