#include <bits/stdc++.h>
using namespace std;
template <class T>
void print(T& x) {
  cout << x;
}
template <class T, class U>
void print(pair<T, U>& p) {
  cout << "(";
  print(p.first);
  cout << ", ";
  print(p.second);
  cout << ")";
}
template <class T>
void print(vector<T>& v) {
  cout << "{";
  if (((int)(v).size())) print(v[0]);
  for (int i = 1; i < ((int)(v).size()); i++) cout << ", ", print(v[i]);
  cout << "}\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int N, M, K;
  long long X, Y;
  cin >> N >> M;
  cin >> X >> K >> Y;
  vector<int> A(N + 1);
  for (int i = 0; i < N; i++) {
    cin >> A[i];
  }
  A[N] = 0;
  vector<int> B(M + 2);
  for (int i = 1; i <= M; i++) {
    cin >> B[i];
  }
  B[M + 1] = 0;
  int ai = 0;
  long long C = 0;
  for (int i = 1; i <= M + 1; i++) {
    int st = ai;
    int m = -1;
    while (ai <= N && A[ai] != B[i]) {
      m = max(m, A[ai]);
      ai++;
    }
    if (ai == N + 1) {
      cout << -1 << endl;
      return 0;
    }
    long long cnt = ai - st;
    if (m > B[i - 1] && m > B[i]) {
      if (cnt < K) {
        cout << -1 << endl;
        return 0;
      }
      C += X;
      cnt -= K;
    }
    if (Y * K <= X) {
      C += Y * cnt;
    } else {
      C += cnt / K * X + cnt % K * Y;
    }
    ai++;
  }
  cout << C << "\n";
}

### Prompt

Construct a cpp code solution to the problem outlined:

There are n warriors in a row. The power of the i-th warrior is a_i. All powers are pairwise distinct.

You have two types of spells which you may cast: 

  1. Fireball: you spend x mana and destroy exactly k consecutive warriors; 
  2. Berserk: you spend y mana, choose two consecutive warriors, and the warrior with greater power destroys the warrior with smaller power. 



For example, let the powers of warriors be [2, 3, 7, 8, 11, 5, 4], and k = 3. If you cast Berserk on warriors with powers 8 and 11, the resulting sequence of powers becomes [2, 3, 7, 11, 5, 4]. Then, for example, if you cast Fireball on consecutive warriors with powers [7, 11, 5], the resulting sequence of powers becomes [2, 3, 4].

You want to turn the current sequence of warriors powers a_1, a_2, ..., a_n into b_1, b_2, ..., b_m. Calculate the minimum amount of mana you need to spend on it.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the length of sequence a and the length of sequence b respectively.

The second line contains three integers x, k, y (1 ≤ x, y, ≤ 10^9; 1 ≤ k ≤ n) — the cost of fireball, the range of fireball and the cost of berserk respectively.

The third line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n). It is guaranteed that all integers a_i are pairwise distinct.

The fourth line contains m integers b_1, b_2, ..., b_m (1 ≤ b_i ≤ n). It is guaranteed that all integers b_i are pairwise distinct.

Output

Print the minimum amount of mana for turning the sequnce a_1, a_2, ..., a_n into b_1, b_2, ..., b_m, or -1 if it is impossible.

Examples

Input


5 2
5 2 3
3 1 4 5 2
3 5


Output


8


Input


4 4
5 1 4
4 3 1 2
2 4 3 1


Output


-1


Input


4 4
2 1 11
1 3 2 4
1 3 2 4


Output


0

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T>
void print(T& x) {
  cout << x;
}
template <class T, class U>
void print(pair<T, U>& p) {
  cout << "(";
  print(p.first);
  cout << ", ";
  print(p.second);
  cout << ")";
}
template <class T>
void print(vector<T>& v) {
  cout << "{";
  if (((int)(v).size())) print(v[0]);
  for (int i = 1; i < ((int)(v).size()); i++) cout << ", ", print(v[i]);
  cout << "}\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int N, M, K;
  long long X, Y;
  cin >> N >> M;
  cin >> X >> K >> Y;
  vector<int> A(N + 1);
  for (int i = 0; i < N; i++) {
    cin >> A[i];
  }
  A[N] = 0;
  vector<int> B(M + 2);
  for (int i = 1; i <= M; i++) {
    cin >> B[i];
  }
  B[M + 1] = 0;
  int ai = 0;
  long long C = 0;
  for (int i = 1; i <= M + 1; i++) {
    int st = ai;
    int m = -1;
    while (ai <= N && A[ai] != B[i]) {
      m = max(m, A[ai]);
      ai++;
    }
    if (ai == N + 1) {
      cout << -1 << endl;
      return 0;
    }
    long long cnt = ai - st;
    if (m > B[i - 1] && m > B[i]) {
      if (cnt < K) {
        cout << -1 << endl;
        return 0;
      }
      C += X;
      cnt -= K;
    }
    if (Y * K <= X) {
      C += Y * cnt;
    } else {
      C += cnt / K * X + cnt % K * Y;
    }
    ai++;
  }
  cout << C << "\n";
}

```