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 long long int N = 5e5 + 5;
const long long int M = 1e6 + 5;
const long long int mod = 1e9 + 7;
const long long int m1 = 1e9 + 7;
const long long int m2 = 1e9 + 9;
const long long int p1 = 402653189;
const long long int p2 = 1610612741;
const int LN = 32;
long long int powermodm(long long int x, long long int n, long long int M) {
  long long int result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % M;
    x = (x * x) % M;
    n = n / 2;
  }
  return result;
}
long long int power(long long int _a, long long int _b) {
  long long int _r = 1;
  while (_b) {
    if (_b % 2 == 1) _r = (_r * _a);
    _b /= 2;
    _a = (_a * _a);
  }
  return _r;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int lcm(long long int a, long long int b) {
  return (max(a, b) / gcd(a, b)) * min(a, b);
}
long long int dx[4] = {-1, 1, 0, 0};
long long int dy[4] = {0, 0, 1, -1};
long long int pw[LN], fact[N], invfact[N];
void pre() {
  fact[0] = 1;
  for (int i = 0; i < LN; i++) pw[i] = power(2, i);
  for (int i = 1; i < N; i++) fact[i] = (i * fact[i - 1]) % mod;
  for (int i = 0; i < N; i++) invfact[i] = powermodm(fact[i], mod - 2, mod);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int i, j;
  long long int n, p, q, r;
  cin >> n >> p >> q >> r;
  long long int a[n + 2], dp_pre[n + 2], dp_suf[n + 2];
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 0; i <= n + 1; i++) {
    dp_pre[i] = dp_suf[i] = -LONG_LONG_MAX;
  }
  for (i = 1; i <= n; i++) {
    dp_pre[i] = max(dp_pre[i - 1], p * a[i]);
  }
  for (i = n; i >= 1; i--) {
    dp_suf[i] = max(dp_suf[i + 1], r * a[i]);
  }
  long long int ans = -LONG_LONG_MAX;
  for (i = 1; i <= n; i++) {
    long long int tans = q * a[i] + dp_pre[i] + dp_suf[i];
    ans = max(ans, tans);
  }
  printf("%lld\n", ans);
  return 0;
}
