Problem:
Score : 1000 points

Problem Statement

You are given a permutation p = (p_1, \ldots, p_N) of \{ 1, \ldots, N \}.
You can perform the following two kinds of operations repeatedly in any order:

Pay a cost A. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the left by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_{l + 1}, p_{l + 2}, \ldots, p_r, p_l, respectively.

Pay a cost B. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the right by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_r, p_l, \ldots, p_{r - 2}, p_{r - 1}, respectively.

Find the minimum total cost required to sort p in ascending order.

Constraints

All values in input are integers.

1 \leq N \leq 5000

1 \leq A, B \leq 10^9

(p_1 \ldots, p_N) is a permutation of \{ 1, \ldots, N \}.

Input

Input is given from Standard Input in the following format:

N A B
p_1 \cdots p_N

Output

Print the minimum total cost required to sort p in ascending order.

Sample Input 1

3 20 30
3 1 2

Sample Output 1

20

Shifting (p_1, p_2, p_3) to the left by one results in p = (1, 2, 3).

Sample Input 2

4 20 30
4 2 3 1

Sample Output 2

50

One possible sequence of operations is as follows:

Shift (p_1, p_2, p_3, p_4) to the left by one. Now we have p = (2, 3, 1, 4).

Shift (p_1, p_2, p_3) to the right by one. Now we have p = (1, 2, 3, 4).

Here, the total cost is 20 + 30 = 50.

Sample Input 3

1 10 10
1

Sample Output 3

0

Sample Input 4

4 1000000000 1000000000
4 3 2 1

Sample Output 4

3000000000

Sample Input 5

9 40 50
5 3 4 7 6 1 2 9 8

Sample Output 5

220

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>

using namespace std;

using int64 = long long;
const int mod = 1e9 + 7;
const int inf = (1 << 30) - 1;
const int64 infll = (1LL << 61) - 1;

struct IoSetup {
  IoSetup() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);
    cerr << fixed << setprecision(10);
  }
} iosetup;

template< typename T >
ostream &operator<<(ostream &os, const vector< T > &v) {
  for(int i = 0; i < (int) v.size(); i++) {
    os << v[i] << (i + 1 != v.size() ? " " : "");
  }
  return os;
}

template< typename T >
istream &operator>>(istream &is, vector< T > &v) {
  for(T &in : v) is >> in;
  return is;
}

template< typename T1, typename T2 >
inline bool chmax(T1 &a, T2 b) { return a < b && (a = b, true); }

template< typename T1, typename T2 >
inline bool chmin(T1 &a, T2 b) { return a > b && (a = b, true); }

template< typename T = int64 >
vector< T > make_v(size_t a) {
  return vector< T >(a);
}

template< typename T, typename... Ts >
auto make_v(size_t a, Ts... ts) {
  return vector< decltype(make_v< T >(ts...)) >(a, make_v< T >(ts...));
}

template< typename T, typename V >
typename enable_if< is_class< T >::value == 0 >::type fill_v(T &t, const V &v) {
  t = v;
}

template< typename T, typename V >
typename enable_if< is_class< T >::value != 0 >::type fill_v(T &t, const V &v) {
  for(auto &e : t) fill_v(e, v);
}

using pi = pair< int, int >;

int main() {
  int N, A, B;
  cin >> N >> A >> B;
  vector< int > V(N);
  cin >> V;

  vector< int > rev(N);
  for(int i = 0; i < N; i++) rev[--V[i]] = i;

  auto dp = make_v< int64 >(N + 1, N + 1);
  fill_v(dp, -1);
  function< int64(int, int) > rec = [&](int idx, int pos) {
    pos = min(pos, N);
    if(idx == N) return 0LL;
    if(~dp[idx][pos]) return dp[idx][pos];
    int64 ret = infll;
    chmin(ret, rec(idx + 1, pos) + B);
    if(rev[idx] < pos) chmin(ret, rec(idx + 1, pos));
    else chmin(ret, rec(idx + 1, (idx - pos) + rev[idx] + 1) + 1LL * (rev[idx] - pos) * A);
    return dp[idx][pos] = ret;
  };
  cout << rec(0, 0) << endl;
}