#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  while (t--) {
    long long int n, m, x, k, y, ans = 0, j = 1;
    cin >> n >> m >> x >> k >> y;
    long long int a[n + 2], b[m + 2], v[m + 2];
    b[0] = 0;
    b[m + 1] = 0;
    v[0] = 0;
    v[m + 1] = n + 1;
    for (long long int i = 1; i <= n; i++) cin >> a[i];
    for (long long int i = 1; i <= m; i++) cin >> b[i];
    for (long long int i = 1; i <= n && j <= m; i++) {
      if (b[j] == a[i]) {
        v[j] = i;
        j++;
      }
    }
    if (j != m + 1) {
      cout << -1;
      return 0;
    }
    for (long long int i = 1; i <= m + 1; i++) {
      long long int len = v[i] - v[i - 1] - 1, st = v[i - 1] + 1,
                    end = v[i] - 1, maxi = 0;
      for (long long int g = st; g <= end; g++) maxi = max(maxi, a[g]);
      if (len < k) {
        if (maxi > max(b[i], b[i - 1])) {
          cout << -1;
          return 0;
        }
        ans += len * y;
      } else {
        if (k * y <= x) {
          if (maxi < max(b[i], b[i - 1]))
            ans += len * y;
          else
            ans += x + (len - k) * y;
        } else
          ans += (len % k) * y + (len / k) * x;
      }
    }
    cout << ans;
  }
}

### Prompt

Generate 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  while (t--) {
    long long int n, m, x, k, y, ans = 0, j = 1;
    cin >> n >> m >> x >> k >> y;
    long long int a[n + 2], b[m + 2], v[m + 2];
    b[0] = 0;
    b[m + 1] = 0;
    v[0] = 0;
    v[m + 1] = n + 1;
    for (long long int i = 1; i <= n; i++) cin >> a[i];
    for (long long int i = 1; i <= m; i++) cin >> b[i];
    for (long long int i = 1; i <= n && j <= m; i++) {
      if (b[j] == a[i]) {
        v[j] = i;
        j++;
      }
    }
    if (j != m + 1) {
      cout << -1;
      return 0;
    }
    for (long long int i = 1; i <= m + 1; i++) {
      long long int len = v[i] - v[i - 1] - 1, st = v[i - 1] + 1,
                    end = v[i] - 1, maxi = 0;
      for (long long int g = st; g <= end; g++) maxi = max(maxi, a[g]);
      if (len < k) {
        if (maxi > max(b[i], b[i - 1])) {
          cout << -1;
          return 0;
        }
        ans += len * y;
      } else {
        if (k * y <= x) {
          if (maxi < max(b[i], b[i - 1]))
            ans += len * y;
          else
            ans += x + (len - k) * y;
        } else
          ans += (len % k) * y + (len / k) * x;
      }
    }
    cout << ans;
  }
}

```