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 int dx[] = {0, 0, -1, 1, 1, -1, -1, 1};
const int dy[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1000000007;
const int N = 1e5 + 50;
int n;
int arr[N], q[3];
long long dp[N][3];
bool vis[N][3];
long long solve(int idx, int p) {
  if (p == 3) return 0;
  if (idx == n) return -(1ll << 63);
  long long &ret = dp[idx][p];
  if (vis[idx][p]) return ret;
  vis[idx][p] = 1;
  if (~ret) return ret;
  return ret =
             max(solve(idx, p + 1) + 1ll * q[p] * arr[idx], solve(idx + 1, p));
}
int main() {
  memset(dp, -1, sizeof(dp));
  cin >> n;
  for (int i = 0; i < 3; ++i) cin >> q[i];
  for (int i = 0; i < n; ++i) cin >> arr[i];
  cout << solve(0, 0) << endl;
  return 0;
}
