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 = 1e5 + 10;
const long long INF = 4e18;
long long a[MAXN];
long long mxx[MAXN], mnn[MAXN];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  for (int i = 0; i < MAXN; i++) {
    mxx[i] = -INF;
    mnn[i] = INF;
  }
  int n;
  long long p, q, r;
  cin >> n >> p >> q >> r;
  long long mn = INF, mx = -INF;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    mn = min(mn, a[i]);
    mx = max(mx, a[i]);
  }
  for (int i = n - 1; i >= 0; i--) {
    mxx[i] = max(mxx[i + 1], a[i]);
    mnn[i] = min(mnn[i + 1], a[i]);
  }
  if (p <= 0 && q <= 0 && r <= 0) {
    cout << mn * (p + q + r) << endl;
  } else if (p >= 0 && q >= 0 && r >= 0) {
    cout << mx * (p + q + r) << endl;
  } else if (p >= 0 && q >= 0 && r <= 0) {
    mx = -INF;
    long long ans = -INF;
    for (int i = 0; i < n; i++) {
      mx = max(mx, a[i]);
      ans = max(ans, mx * (p + q) + r * a[i]);
    }
    cout << ans << endl;
  } else if (p >= 0 && q <= 0 && r <= 0) {
    mx = -INF;
    long long ans = -INF;
    for (int i = 0; i < n; i++) {
      mx = max(mx, a[i]);
      ans = max(ans, mx * p + (q + r) * a[i]);
    }
    cout << ans << endl;
  } else if (p <= 0 && q <= 0 && r >= 0) {
    mn = INF;
    long long ans = -INF;
    for (int i = 0; i < n; i++) {
      mn = min(mn, a[i]);
      ans = max(ans, mn * (p + q) + r * a[i]);
    }
    cout << ans << endl;
  } else if (p <= 0 && q >= 0 && r >= 0) {
    mn = INF;
    long long ans = -INF;
    for (int i = 0; i < n; i++) {
      mn = min(mn, a[i]);
      ans = max(ans, mn * p + (q + r) * a[i]);
    }
    cout << ans << endl;
  } else if (p <= 0 && q >= 0 && r <= 0) {
    mn = INF;
    mx = -INF;
    long long ans = -INF;
    for (int i = 0; i < n; i++) {
      mn = min(mn, a[i]);
      ans = max(ans, mn * p + q * a[i] + r * mnn[i]);
    }
    cout << ans << endl;
  } else {
    mn = INF;
    mx = -INF;
    long long ans = -INF;
    for (int i = 0; i < n; i++) {
      mx = max(mx, a[i]);
      ans = max(ans, mx * p + q * a[i] + r * mxx[i]);
    }
    cout << ans << endl;
  }
  return 0;
}
