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;
void redwackyferret() {
  ios::sync_with_stdio(0);
  cin.tie(0);
}
int32_t main() {
  redwackyferret();
  int p, q, r, n;
  cin >> n >> p >> q >> r;
  vector<int> nums(n);
  for (int i = 0; i < n; i++) cin >> nums[i];
  long long dp[n][3];
  dp[0][0] = 1LL * p * nums[0];
  for (int i = 1; i < n; i++) {
    dp[i][0] = max(dp[i - 1][0], 1LL * p * nums[i]);
  }
  dp[0][1] = dp[0][0] + 1ll * q * nums[0];
  for (int i = 1; i < n; i++) {
    dp[i][1] = max(dp[i - 1][1], dp[i][0] + 1ll * q * nums[i]);
  }
  dp[0][2] = dp[0][1] + 1ll * r * nums[0];
  for (int i = 1; i < n; i++) {
    dp[i][2] = max(dp[i - 1][2], dp[i][1] + 1ll * r * nums[i]);
  }
  cout << dp[n - 1][2] << endl;
  return 0;
}
