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 MAXN = 100100;
const int MOD7 = 1000000007;
const int MOD9 = 1000000009;
const int INF = 2000000000;
const double EPS = 1e-9;
const double PI = 3.14159265358979;
const int dir_4r[] = {-1, 1, 0, 0};
const int dir_4c[] = {0, 0, -1, 1};
const int dir_8r[] = {-1, -1, -1, 0, 0, 1, 1, 1};
const int dir_8c[] = {-1, 0, 1, -1, 1, -1, 0, 1};
int val[MAXN];
long long dp[MAXN][3];
int main() {
  int n, p, q, r;
  scanf("%d%d%d%d", &n, &p, &q, &r);
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i < n; ++i) scanf("%d", val + i);
  dp[0][0] = (long long)val[0] * p;
  for (int i = 1; i < n; ++i)
    dp[i][0] = max(dp[i - 1][0], (long long)val[i] * p);
  dp[0][1] = dp[0][0] + (long long)val[0] * q;
  for (int i = 1; i < n; ++i)
    dp[i][1] = max(dp[i - 1][1], dp[i][0] + (long long)val[i] * q);
  dp[0][2] = dp[0][1] + (long long)val[0] * r;
  for (int i = 1; i < n; ++i)
    dp[i][2] = max(dp[i - 1][2], dp[i][1] + (long long)val[i] * r);
  printf("%I64d\n", dp[n - 1][2]);
  return 0;
}
