#include <bits/stdc++.h>
using namespace std;
long long a[200005];
long long b[200005];
int pos[200005];
int goal[200005];
int main() {
  int n, m;
  cin >> n >> m;
  long long cf, cb;
  int k;
  cin >> cf >> k >> cb;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    a[i] = x;
    pos[x] = i;
  }
  bool can = true;
  goal[1] = 0;
  goal[m + 2] = n + 1;
  for (int i = 1; i <= m; i++) {
    int x;
    cin >> x;
    goal[i + 1] = pos[x];
    b[i] = x;
    if (goal[i + 1] == 0 || (goal[i] > goal[i + 1])) {
      can = false;
    }
  }
  if (!can) {
    cout << -1 << '\n';
    return 0;
  }
  long long res = 0;
  for (int i = 2; i <= m + 2 && can; i++) {
    int diff = goal[i] - goal[i - 1] - 1;
    if (diff > 0) {
      bool more = true;
      for (int j = goal[i - 1] + 1; j <= goal[i] - 1; j++) {
        if (a[j] > max(a[goal[i]], a[goal[i - 1]])) {
          more = false;
        }
      }
      if (more) {
        int left = diff;
        res += cb * (left % k);
        left -= (left % k);
        int possible = left / k;
        if (k * cb > cf) {
          res += cf * possible;
        } else {
          res += cb * k * possible;
        }
      } else {
        if (diff < k) {
          can = false;
        } else {
          int left = diff - k;
          res += cb * (left % k);
          left -= (left % k);
          int possible = left / k;
          if (k * cb > cf) {
            res += cf * possible;
          } else {
            res += cb * k * possible;
          }
          res += cf;
        }
      }
    }
  }
  if (!can) {
    cout << -1 << '\n';
    return 0;
  }
  cout << res << '\n';
}

### Prompt

Please formulate a Cpp solution to 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 a[200005];
long long b[200005];
int pos[200005];
int goal[200005];
int main() {
  int n, m;
  cin >> n >> m;
  long long cf, cb;
  int k;
  cin >> cf >> k >> cb;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    a[i] = x;
    pos[x] = i;
  }
  bool can = true;
  goal[1] = 0;
  goal[m + 2] = n + 1;
  for (int i = 1; i <= m; i++) {
    int x;
    cin >> x;
    goal[i + 1] = pos[x];
    b[i] = x;
    if (goal[i + 1] == 0 || (goal[i] > goal[i + 1])) {
      can = false;
    }
  }
  if (!can) {
    cout << -1 << '\n';
    return 0;
  }
  long long res = 0;
  for (int i = 2; i <= m + 2 && can; i++) {
    int diff = goal[i] - goal[i - 1] - 1;
    if (diff > 0) {
      bool more = true;
      for (int j = goal[i - 1] + 1; j <= goal[i] - 1; j++) {
        if (a[j] > max(a[goal[i]], a[goal[i - 1]])) {
          more = false;
        }
      }
      if (more) {
        int left = diff;
        res += cb * (left % k);
        left -= (left % k);
        int possible = left / k;
        if (k * cb > cf) {
          res += cf * possible;
        } else {
          res += cb * k * possible;
        }
      } else {
        if (diff < k) {
          can = false;
        } else {
          int left = diff - k;
          res += cb * (left % k);
          left -= (left % k);
          int possible = left / k;
          if (k * cb > cf) {
            res += cf * possible;
          } else {
            res += cb * k * possible;
          }
          res += cf;
        }
      }
    }
  }
  if (!can) {
    cout << -1 << '\n';
    return 0;
  }
  cout << res << '\n';
}

```