#include <bits/stdc++.h>
using namespace std;
long long n, m, x, k, y, a[200005], b[200005];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> x >> k >> y;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < m; ++i) {
    cin >> b[i];
  }
  vector<long long> contain;
  contain.push_back(-1);
  int index = 0;
  for (int i = 0; i < n && index < m; ++i) {
    if (a[i] == b[index]) {
      contain.push_back(i);
      index++;
    }
  }
  contain.push_back(n);
  if (index < m) {
    cout << -1 << "\n";
    return 0;
  }
  long long res = 0;
  if (x <= k * y) {
    for (int i = 1; i < contain.size(); ++i) {
      long long counter = 0;
      for (int j = contain[i - 1] + 1; j < contain[i]; ++j) {
        if ((contain[i] == n || a[j] > a[contain[i]]) &&
            (contain[i - 1] == -1 || a[j] > a[contain[i - 1]])) {
          if (contain[i] - j >= k) {
            res += x;
            j += (k - 1);
          } else if (contain[i] - j + counter >= k) {
            res += x;
            counter -= (k - contain[i] + j);
            break;
          } else {
            cout << -1 << "\n";
            return 0;
          }
        } else {
          counter++;
        }
      }
      res += (counter / k) * x;
      res += (counter % k) * y;
    }
  } else {
    for (int i = 1; i < contain.size(); ++i) {
      bool large = false;
      for (int j = contain[i - 1] + 1; j < contain[i]; ++j) {
        if ((contain[i] == n || a[j] > a[contain[i]]) &&
            (contain[i - 1] == -1 || a[j] > a[contain[i - 1]])) {
          large = true;
          break;
        }
      }
      long long size = contain[i] - contain[i - 1] - 1;
      if (large) {
        if (size >= k) {
          res += x;
          size -= k;
        } else {
          cout << -1 << "\n";
          return 0;
        }
      }
      res += size * y;
    }
  }
  cout << res << "\n";
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 n, m, x, k, y, a[200005], b[200005];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> x >> k >> y;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < m; ++i) {
    cin >> b[i];
  }
  vector<long long> contain;
  contain.push_back(-1);
  int index = 0;
  for (int i = 0; i < n && index < m; ++i) {
    if (a[i] == b[index]) {
      contain.push_back(i);
      index++;
    }
  }
  contain.push_back(n);
  if (index < m) {
    cout << -1 << "\n";
    return 0;
  }
  long long res = 0;
  if (x <= k * y) {
    for (int i = 1; i < contain.size(); ++i) {
      long long counter = 0;
      for (int j = contain[i - 1] + 1; j < contain[i]; ++j) {
        if ((contain[i] == n || a[j] > a[contain[i]]) &&
            (contain[i - 1] == -1 || a[j] > a[contain[i - 1]])) {
          if (contain[i] - j >= k) {
            res += x;
            j += (k - 1);
          } else if (contain[i] - j + counter >= k) {
            res += x;
            counter -= (k - contain[i] + j);
            break;
          } else {
            cout << -1 << "\n";
            return 0;
          }
        } else {
          counter++;
        }
      }
      res += (counter / k) * x;
      res += (counter % k) * y;
    }
  } else {
    for (int i = 1; i < contain.size(); ++i) {
      bool large = false;
      for (int j = contain[i - 1] + 1; j < contain[i]; ++j) {
        if ((contain[i] == n || a[j] > a[contain[i]]) &&
            (contain[i - 1] == -1 || a[j] > a[contain[i - 1]])) {
          large = true;
          break;
        }
      }
      long long size = contain[i] - contain[i - 1] - 1;
      if (large) {
        if (size >= k) {
          res += x;
          size -= k;
        } else {
          cout << -1 << "\n";
          return 0;
        }
      }
      res += size * y;
    }
  }
  cout << res << "\n";
  return 0;
}

```