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;
inline long long isqrt(long long k) {
  long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline long long minize(long long& a, long long b) {
  return a = (a > b ? b : a);
}
inline long long maxize(long long& a, long long b) {
  return a = (a < b ? b : a);
}
inline string toString(long long n) {
  stringstream ss;
  ss << n;
  return ss.str();
}
double const eps = 1e-6;
long long const Base = 1e9 + 7, oo = 1e17, MAXN = 1e5;
long long arr[MAXN + 5], arr2[5];
pair<long long, long long> dp[5][MAXN + 5];
long long Solves() {
  long long k, cnt = 0, ans = 0, x, y, q, sum = 0, v, t, n, a, b, c,
               maxval = -oo, minval = oo, p1, p2, p3;
  cin >> n;
  cin >> arr2[1] >> arr2[2] >> arr2[3];
  for (long long i = 1; i <= n; i++) cin >> arr[i];
  for (long long i = 1; i <= 3; i++) {
    for (long long j = 1; j <= n; j++) {
      dp[i][j].first = min(arr2[i] * arr[j] + dp[i - 1][j].first,
                           arr2[i] * arr[j] + dp[i - 1][j].second);
      if (j > 1) minize(dp[i][j].first, dp[i][j - 1].first);
      dp[i][j].second = max(arr2[i] * arr[j] + dp[i - 1][j].first,
                            arr2[i] * arr[j] + dp[i - 1][j].second);
      if (j > 1) maxize(dp[i][j].second, dp[i][j - 1].second);
    }
  }
  cout << dp[3][n].second;
  return 0;
}
int main() {
  long long JUDGE_ONLINE = 1;
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  Solves();
  if (!JUDGE_ONLINE)
    cout << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
}
