#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m, x, k, y;
  cin >> n >> m >> x >> k >> y;
  vector<int> a(n + 2), b(m);
  a[0] = a[n + 1] = -1;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < m; ++i) {
    cin >> b[i];
  }
  vector<int> idx = {0};
  for (int i = 1, j = 0; i <= n && j < m;) {
    if (a[i] == b[j]) {
      idx.push_back(i);
      ++i, ++j;
    } else {
      ++i;
    }
  }
  idx.push_back(n + 1);
  if (idx.size() != m + 2) {
    cout << -1 << '\n';
    return 0;
  }
  long long an = 0;
  for (int i = 1; i < m + 2; ++i) {
    int big = 0, prev = idx[i - 1], curr = idx[i], cnt = curr - prev - 1;
    for (int j = prev + 1; j < curr; ++j) {
      if (a[prev] < a[j] && a[curr] < a[j]) big++;
    }
    if (big > 0 && cnt < k) {
      cout << -1 << '\n';
      return 0;
    }
    an += (cnt % k) * y;
    cnt -= (cnt % k);
    if (x <= y * k) {
      an += 1ll * cnt / k * x;
    } else if (big > 0) {
      an += 1ll * x + 1ll * (cnt - k) * y;
    } else {
      an += 1ll * cnt * y;
    }
  }
  cout << an << '\n';
  return 0;
}

### 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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m, x, k, y;
  cin >> n >> m >> x >> k >> y;
  vector<int> a(n + 2), b(m);
  a[0] = a[n + 1] = -1;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < m; ++i) {
    cin >> b[i];
  }
  vector<int> idx = {0};
  for (int i = 1, j = 0; i <= n && j < m;) {
    if (a[i] == b[j]) {
      idx.push_back(i);
      ++i, ++j;
    } else {
      ++i;
    }
  }
  idx.push_back(n + 1);
  if (idx.size() != m + 2) {
    cout << -1 << '\n';
    return 0;
  }
  long long an = 0;
  for (int i = 1; i < m + 2; ++i) {
    int big = 0, prev = idx[i - 1], curr = idx[i], cnt = curr - prev - 1;
    for (int j = prev + 1; j < curr; ++j) {
      if (a[prev] < a[j] && a[curr] < a[j]) big++;
    }
    if (big > 0 && cnt < k) {
      cout << -1 << '\n';
      return 0;
    }
    an += (cnt % k) * y;
    cnt -= (cnt % k);
    if (x <= y * k) {
      an += 1ll * cnt / k * x;
    } else if (big > 0) {
      an += 1ll * x + 1ll * (cnt - k) * y;
    } else {
      an += 1ll * cnt * y;
    }
  }
  cout << an << '\n';
  return 0;
}

```