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>
long long int mod = 1000000007;
using namespace std;
long long int n, p, q, r;
long long int a[100000];
long long int dp[100000][3];
bool vis[100001][3];
bool calc(int x, int y) {
  if (vis[x][y]) {
    return vis[x][y];
  }
  if (x == n - 1) {
    if (y == 0) {
      dp[x][y] = (p + q + r) * a[x];
    } else if (y == 1) {
      dp[x][y] = (q + r) * (a[x]);
    } else {
      dp[x][y] = r * a[x];
    }
    vis[x][y] = 1;
    return vis[x][y];
  }
  if (y == 2) {
    dp[x][y] = r * a[x];
    vis[x][y] = 1;
    calc(x + 1, y);
    dp[x][y] = max(dp[x][y], dp[x + 1][y]);
    return vis[x][y];
  }
  if (y == 1) {
    vis[x][y] = 1;
    calc(x, y + 1);
    dp[x][y] = q * a[x] + dp[x][y + 1];
    calc(x + 1, y);
    dp[x][y] = max(dp[x][y], dp[x + 1][y]);
    return vis[x][y];
  }
  if (y == 0) {
    calc(x, y + 1);
    dp[x][y] = p * a[x] + dp[x][y + 1];
    calc(x + 1, y);
    dp[x][y] = max(dp[x][y], dp[x + 1][y]);
    vis[x][y] = 1;
    return vis[x][y];
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  cin >> n >> p >> q >> r;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  memset(vis, false, sizeof(vis));
  calc(0, 0);
  cout << dp[0][0] << endl;
  return 0;
}
