#include <bits/stdc++.h>
using namespace std;
long long a[200100];
long long b[200100];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m, x, k, y;
  cin >> n >> m;
  cin >> x >> k >> y;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= m; i++) cin >> b[i];
  long long p = 1;
  long long flag = 0;
  if (y * k > x) flag = 1;
  long long swi = 1, sta = 0, maxn = 0;
  long long cost = 0;
  long long ans = 1;
  for (int i = 1; i <= n; i++) {
    if (a[i] > a[sta]) {
      if (p <= m && a[i] > b[p])
        maxn++;
      else if (p > m)
        maxn++;
    }
    if (a[i] == b[p] || i == n) {
      long long people = i - sta - 1;
      if (a[i] != b[p]) people++;
      p++;
      sta = i;
      if (maxn > 0) {
        long long left = people % k;
        if (people < k) {
          ans = 0;
          break;
        }
        if (flag) {
          cost += x * (people / k);
          cost += left * y;
        } else {
          cost += x;
          cost += (people - k) * y;
        }
        maxn = 0;
      } else {
        long long left = people % k;
        if (flag) {
          cost += x * (people / k);
          cost += left * y;
        } else {
          cost += y * people;
        }
      }
    }
  }
  if (p <= m || ans == 0)
    cout << "-1\n";
  else
    cout << cost << "\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 a[200100];
long long b[200100];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m, x, k, y;
  cin >> n >> m;
  cin >> x >> k >> y;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= m; i++) cin >> b[i];
  long long p = 1;
  long long flag = 0;
  if (y * k > x) flag = 1;
  long long swi = 1, sta = 0, maxn = 0;
  long long cost = 0;
  long long ans = 1;
  for (int i = 1; i <= n; i++) {
    if (a[i] > a[sta]) {
      if (p <= m && a[i] > b[p])
        maxn++;
      else if (p > m)
        maxn++;
    }
    if (a[i] == b[p] || i == n) {
      long long people = i - sta - 1;
      if (a[i] != b[p]) people++;
      p++;
      sta = i;
      if (maxn > 0) {
        long long left = people % k;
        if (people < k) {
          ans = 0;
          break;
        }
        if (flag) {
          cost += x * (people / k);
          cost += left * y;
        } else {
          cost += x;
          cost += (people - k) * y;
        }
        maxn = 0;
      } else {
        long long left = people % k;
        if (flag) {
          cost += x * (people / k);
          cost += left * y;
        } else {
          cost += y * people;
        }
      }
    }
  }
  if (p <= m || ans == 0)
    cout << "-1\n";
  else
    cout << cost << "\n";
  return 0;
}

```