#include <bits/stdc++.h>
using namespace std;
long long n, m, x, k, y;
long long a[200005], b[200005];
long long pos[200005];
int main() {
  cin >> n >> m;
  cin >> x >> k >> y;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    pos[a[i]] = i;
  }
  for (int i = 1; i <= m; i++) {
    cin >> b[i];
  }
  if (m > n) {
    cout << -1;
    return 0;
  }
  for (int i = 2; i <= m; i++) {
    if (pos[b[i]] < pos[b[i - 1]]) {
      cout << -1;
      return 0;
    }
  }
  long long ans = 0;
  for (int i = 1; i <= m + 1; i++) {
    if (i == m + 1) pos[0] = n + 1;
    int dif = pos[b[i]] - pos[b[i - 1]] - 1;
    if (dif == 0) continue;
    long long maxim = 0;
    for (int j = pos[b[i - 1]] + 1; j < pos[b[i]]; j++) {
      maxim = max(maxim, a[j]);
    }
    if (max(b[i], b[i - 1]) > maxim) {
      long long cx = min(x, y * k);
      ans += cx * (dif / k) + y * (dif % k);
    } else {
      if (dif < k) {
        cout << -1;
        return 0;
      } else {
        if (y * k < x) {
          ans += (dif - k) * y + x;
        } else {
          long long cx = min(x, y * k);
          ans += cx * (dif / k) + y * (dif % k);
        }
      }
    }
  }
  cout << ans;
}

### 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 n, m, x, k, y;
long long a[200005], b[200005];
long long pos[200005];
int main() {
  cin >> n >> m;
  cin >> x >> k >> y;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    pos[a[i]] = i;
  }
  for (int i = 1; i <= m; i++) {
    cin >> b[i];
  }
  if (m > n) {
    cout << -1;
    return 0;
  }
  for (int i = 2; i <= m; i++) {
    if (pos[b[i]] < pos[b[i - 1]]) {
      cout << -1;
      return 0;
    }
  }
  long long ans = 0;
  for (int i = 1; i <= m + 1; i++) {
    if (i == m + 1) pos[0] = n + 1;
    int dif = pos[b[i]] - pos[b[i - 1]] - 1;
    if (dif == 0) continue;
    long long maxim = 0;
    for (int j = pos[b[i - 1]] + 1; j < pos[b[i]]; j++) {
      maxim = max(maxim, a[j]);
    }
    if (max(b[i], b[i - 1]) > maxim) {
      long long cx = min(x, y * k);
      ans += cx * (dif / k) + y * (dif % k);
    } else {
      if (dif < k) {
        cout << -1;
        return 0;
      } else {
        if (y * k < x) {
          ans += (dif - k) * y + x;
        } else {
          long long cx = min(x, y * k);
          ans += cx * (dif / k) + y * (dif % k);
        }
      }
    }
  }
  cout << ans;
}

```