#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, m, x, k, y;
  cin >> n >> m >> x >> k >> y;
  long long int a[n + 2];
  long long int b[m + 2];
  map<long long int, long long int> mp;
  a[0] = 0;
  a[n + 1] = n + 1;
  for (long long int i = 1; i <= n; i++) {
    cin >> a[i];
    mp[a[i]] = i;
  }
  mp[0] = 0;
  mp[n + 1] = n + 1;
  b[0] = 0;
  b[m + 1] = n + 1;
  for (long long int i = 1; i <= m; i++) {
    cin >> b[i];
  }
  bool flag = true;
  long long int ans = 0;
  for (long long int i = 1; i <= m + 1; i++) {
    long long int p = mp[b[i - 1]];
    long long int q = mp[b[i]];
    long long int diff = q - p;
    if (diff > 0) {
      long long int maxm = LLONG_MIN;
      for (long long int j = p + 1; j < q; j++) {
        maxm = max(a[j], maxm);
      }
      long long int dist = q - p - 1;
      if (dist >= k) {
        if (x < k * y) {
          ans += (dist / k) * x + (dist % k) * y;
        } else if (p != 0 && q != n + 1 && maxm < max(a[p], a[q])) {
          ans += (dist)*y;
        } else if (p == 0 && maxm < a[q]) {
          ans += (dist)*y;
        } else if (q == n + 1 && maxm < a[p]) {
          ans += (dist)*y;
        } else {
          ans += x + (dist - k) * y;
        }
      } else {
        if (p != 0 && q != n + 1 && maxm < max(a[p], a[q])) {
          ans += (dist)*y;
        } else if (p == 0 && maxm < a[q]) {
          ans += (dist)*y;
        } else if (q == n + 1 && maxm < a[p]) {
          ans += (dist)*y;
        } else {
          flag = false;
        }
      }
    } else {
      flag = false;
      break;
    }
    if (!flag) {
      break;
    }
  }
  if (flag) {
    cout << ans << endl;
  } else {
    cout << "-1" << endl;
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, m, x, k, y;
  cin >> n >> m >> x >> k >> y;
  long long int a[n + 2];
  long long int b[m + 2];
  map<long long int, long long int> mp;
  a[0] = 0;
  a[n + 1] = n + 1;
  for (long long int i = 1; i <= n; i++) {
    cin >> a[i];
    mp[a[i]] = i;
  }
  mp[0] = 0;
  mp[n + 1] = n + 1;
  b[0] = 0;
  b[m + 1] = n + 1;
  for (long long int i = 1; i <= m; i++) {
    cin >> b[i];
  }
  bool flag = true;
  long long int ans = 0;
  for (long long int i = 1; i <= m + 1; i++) {
    long long int p = mp[b[i - 1]];
    long long int q = mp[b[i]];
    long long int diff = q - p;
    if (diff > 0) {
      long long int maxm = LLONG_MIN;
      for (long long int j = p + 1; j < q; j++) {
        maxm = max(a[j], maxm);
      }
      long long int dist = q - p - 1;
      if (dist >= k) {
        if (x < k * y) {
          ans += (dist / k) * x + (dist % k) * y;
        } else if (p != 0 && q != n + 1 && maxm < max(a[p], a[q])) {
          ans += (dist)*y;
        } else if (p == 0 && maxm < a[q]) {
          ans += (dist)*y;
        } else if (q == n + 1 && maxm < a[p]) {
          ans += (dist)*y;
        } else {
          ans += x + (dist - k) * y;
        }
      } else {
        if (p != 0 && q != n + 1 && maxm < max(a[p], a[q])) {
          ans += (dist)*y;
        } else if (p == 0 && maxm < a[q]) {
          ans += (dist)*y;
        } else if (q == n + 1 && maxm < a[p]) {
          ans += (dist)*y;
        } else {
          flag = false;
        }
      }
    } else {
      flag = false;
      break;
    }
    if (!flag) {
      break;
    }
  }
  if (flag) {
    cout << ans << endl;
  } else {
    cout << "-1" << endl;
  }
  return 0;
}

```