#include <bits/stdc++.h>
using namespace std;
long long int n, m, x, k, y;
std::vector<int> a;
std::vector<int> b;
long long int findans(int lbound, int rbound) {
  if (rbound - lbound == 1) return 0;
  long long int best = INT_MAX;
  int mx = *max_element(a.begin() + lbound + 1, a.begin() + rbound);
  long long int len = rbound - lbound - 1;
  if (len < k) {
    if (mx > max(a[lbound], a[rbound]))
      return -1;
    else {
      return len * y;
    }
  } else {
    if (k * y >= x) {
      long long int ans = (len % k) * y + (len / k) * x;
      return ans;
    } else {
      if (mx < max(a[lbound], a[rbound])) {
        return len * y;
      } else {
        return (x + (len - k) * y);
      }
    }
  }
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  cin >> x >> k >> y;
  a.resize(n + 1);
  b.resize(m);
  a[0] = INT_MIN;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i + 1];
  }
  a.push_back(INT_MIN);
  for (long long int i = 0; i < m; i++) {
    cin >> b[i];
  }
  int j = 0;
  int i;
  int prev = 0;
  long long int ans = 0;
  for (i = 1; i <= n; ++i) {
    if (a[i] == b[j]) {
      long long int temp = findans(prev, i);
      if (temp == -1) {
        cout << "-1\n";
        return 0;
      }
      ans += temp;
      prev = i;
      j++;
    }
  }
  if (j != m) {
    cout << "-1\n";
    return 0;
  }
  long long int temp = findans(prev, n + 1);
  if (temp == -1) {
    cout << "-1\n";
    return 0;
  }
  ans += temp;
  ;
  cout << ans << "\n";
}

### Prompt

Your challenge is to write 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;
long long int n, m, x, k, y;
std::vector<int> a;
std::vector<int> b;
long long int findans(int lbound, int rbound) {
  if (rbound - lbound == 1) return 0;
  long long int best = INT_MAX;
  int mx = *max_element(a.begin() + lbound + 1, a.begin() + rbound);
  long long int len = rbound - lbound - 1;
  if (len < k) {
    if (mx > max(a[lbound], a[rbound]))
      return -1;
    else {
      return len * y;
    }
  } else {
    if (k * y >= x) {
      long long int ans = (len % k) * y + (len / k) * x;
      return ans;
    } else {
      if (mx < max(a[lbound], a[rbound])) {
        return len * y;
      } else {
        return (x + (len - k) * y);
      }
    }
  }
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  cin >> x >> k >> y;
  a.resize(n + 1);
  b.resize(m);
  a[0] = INT_MIN;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i + 1];
  }
  a.push_back(INT_MIN);
  for (long long int i = 0; i < m; i++) {
    cin >> b[i];
  }
  int j = 0;
  int i;
  int prev = 0;
  long long int ans = 0;
  for (i = 1; i <= n; ++i) {
    if (a[i] == b[j]) {
      long long int temp = findans(prev, i);
      if (temp == -1) {
        cout << "-1\n";
        return 0;
      }
      ans += temp;
      prev = i;
      j++;
    }
  }
  if (j != m) {
    cout << "-1\n";
    return 0;
  }
  long long int temp = findans(prev, n + 1);
  if (temp == -1) {
    cout << "-1\n";
    return 0;
  }
  ans += temp;
  ;
  cout << ans << "\n";
}

```