#include <bits/stdc++.h>
using namespace std;
long long x, y, k;
long long solve(vector<long long>& A) {
  bool ok = 1;
  for (int i = 1; i < A.size(); i++) {
    if (A[i] > A[0] && A[i] > A.back()) {
      ok = 0;
    }
  }
  if (A.size() - 2 < k) {
    if (ok == 0) {
      cout << -1;
      exit(0);
    }
    return (y * (A.size() - 2));
  }
  int cnt = A.size() - 2;
  if (x <= y * k) {
    return x * ((cnt) / k) + (y * (cnt % k));
  }
  if (ok) {
    return y * cnt;
  }
  return (x + y * (cnt - k));
}
int main() {
  int n, m;
  cin >> n >> m;
  cin >> x >> k >> y;
  long long D[n], F[m];
  bool used[n];
  int index[n];
  for (int i = 0; i < n; ++i) {
    cin >> D[i];
    D[i]--;
    used[i] = 0;
    index[D[i]] = i;
  }
  int ind = -1;
  for (int j = 0; j < m; ++j) {
    cin >> F[j];
    F[j]--;
    used[F[j]] = 1;
    if (index[F[j]] < ind) {
      cout << -1 << "\n";
      return 0;
    }
    ind = max(ind, index[F[j]]);
  }
  vector<vector<long long> > G;
  vector<long long> t;
  t.push_back(-1);
  for (int i = 0; i < n; ++i) {
    if (used[D[i]] == 0) {
      t.push_back(D[i]);
    } else {
      t.push_back(D[i]);
      if (t.size() > 2) {
        G.push_back(t);
      }
      t.clear();
      t.push_back(D[i]);
    }
  }
  t.push_back(-1);
  if (t.size() > 2) {
    G.push_back(t);
  }
  long long ans = 0;
  for (int i = 0; i < G.size(); ++i) {
    ans = ans + solve(G[i]);
  }
  cout << ans;
  return 0;
}

### Prompt

In Cpp, your task is to solve the following problem:

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;
long long x, y, k;
long long solve(vector<long long>& A) {
  bool ok = 1;
  for (int i = 1; i < A.size(); i++) {
    if (A[i] > A[0] && A[i] > A.back()) {
      ok = 0;
    }
  }
  if (A.size() - 2 < k) {
    if (ok == 0) {
      cout << -1;
      exit(0);
    }
    return (y * (A.size() - 2));
  }
  int cnt = A.size() - 2;
  if (x <= y * k) {
    return x * ((cnt) / k) + (y * (cnt % k));
  }
  if (ok) {
    return y * cnt;
  }
  return (x + y * (cnt - k));
}
int main() {
  int n, m;
  cin >> n >> m;
  cin >> x >> k >> y;
  long long D[n], F[m];
  bool used[n];
  int index[n];
  for (int i = 0; i < n; ++i) {
    cin >> D[i];
    D[i]--;
    used[i] = 0;
    index[D[i]] = i;
  }
  int ind = -1;
  for (int j = 0; j < m; ++j) {
    cin >> F[j];
    F[j]--;
    used[F[j]] = 1;
    if (index[F[j]] < ind) {
      cout << -1 << "\n";
      return 0;
    }
    ind = max(ind, index[F[j]]);
  }
  vector<vector<long long> > G;
  vector<long long> t;
  t.push_back(-1);
  for (int i = 0; i < n; ++i) {
    if (used[D[i]] == 0) {
      t.push_back(D[i]);
    } else {
      t.push_back(D[i]);
      if (t.size() > 2) {
        G.push_back(t);
      }
      t.clear();
      t.push_back(D[i]);
    }
  }
  t.push_back(-1);
  if (t.size() > 2) {
    G.push_back(t);
  }
  long long ans = 0;
  for (int i = 0; i < G.size(); ++i) {
    ans = ans + solve(G[i]);
  }
  cout << ans;
  return 0;
}

```