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 = 100005;
int a[maxn];
int dpmax[maxn][40], dpmin[maxn][40];
int n;
long long p, q, r;
void makeMax() {
  for (int i = 1; i <= n; i++) dpmax[i][0] = a[i];
  for (int j = 1; j <= log((double)n) / log(2.0); j++) {
    for (int i = 1; i + (1 << j) - 1 <= n; i++) {
      dpmax[i][j] = max(dpmax[i][j - 1], dpmax[i + (1 << (j - 1))][j - 1]);
    }
  }
}
void makeMin() {
  for (int i = 1; i <= n; i++) dpmin[i][0] = a[i];
  for (int j = 1; j <= log((double)n) / log(2.0); j++) {
    for (int i = 1; i + (1 << j) - 1 <= n; i++) {
      dpmin[i][j] = min(dpmin[i][j - 1], dpmin[i + (1 << (j - 1))][j - 1]);
    }
  }
}
int getMax(int l, int r) {
  int k = (int)(log((double)(r - l + 1)) / log(2.0));
  return max(dpmax[l][k], dpmax[r - (1 << k) + 1][k]);
}
int getMin(int l, int r) {
  int k = (int)(log((double)(r - l + 1)) / log(2.0));
  return min(dpmin[l][k], dpmin[r - (1 << k) + 1][k]);
}
int main() {
  cin >> n >> p >> q >> r;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  makeMax(), makeMin();
  long long ans = -0x3f3f3f3f3f3f3f3f;
  for (int i = 1; i <= n; i++) {
    long long t = q * (long long)a[i];
    if (p > 0) {
      t += p * (long long)getMax(1, i);
    } else {
      t += p * (long long)getMin(1, i);
    }
    if (r > 0) {
      t += r * (long long)getMax(i, n);
    } else {
      t += r * (long long)getMin(i, n);
    }
    ans = max(ans, t);
  }
  cout << ans << endl;
  return 0;
}
