#include <bits/stdc++.h>
double PI = 3.1415926535;
const int N = (int)(1 * 1e6 + 10);
using namespace std;
int main() {
  int n, m;
  cin >> n >> m;
  long long int x, k, y;
  cin >> x >> k >> y;
  vector<int> a(n + 1);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  a[n] = 0;
  vector<int> b(m + 1);
  for (int i = 0; i < m; i++) {
    cin >> b[i];
  }
  b[m] = 0;
  int j = 0;
  int prev = 0;
  int maxi = 0;
  int cnt = 0;
  long long int result = 0;
  for (int i = 0; i <= n; i++) {
    if (a[i] == b[j]) {
      j++;
      if (cnt < k) {
        if (maxi < prev || maxi < a[i]) {
          result += (cnt * y);
        } else {
          cout << "-1" << endl;
          return 0;
        }
      } else {
        if (maxi < prev || maxi < a[i]) {
          int var = cnt / k;
          result += min(var * x, var * k * y);
          result += y * (cnt % k);
        } else {
          result += x;
          cnt = cnt - k;
          int var = cnt / k;
          result += min(var * x, var * k * y);
          result += y * (cnt % k);
        }
      }
      maxi = 0;
      prev = a[i];
      cnt = 0;
    } else {
      cnt++;
      maxi = max(maxi, a[i]);
    }
  }
  if (j < m) {
    cout << "-1" << endl;
    return 0;
  }
  cout << result << endl;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
double PI = 3.1415926535;
const int N = (int)(1 * 1e6 + 10);
using namespace std;
int main() {
  int n, m;
  cin >> n >> m;
  long long int x, k, y;
  cin >> x >> k >> y;
  vector<int> a(n + 1);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  a[n] = 0;
  vector<int> b(m + 1);
  for (int i = 0; i < m; i++) {
    cin >> b[i];
  }
  b[m] = 0;
  int j = 0;
  int prev = 0;
  int maxi = 0;
  int cnt = 0;
  long long int result = 0;
  for (int i = 0; i <= n; i++) {
    if (a[i] == b[j]) {
      j++;
      if (cnt < k) {
        if (maxi < prev || maxi < a[i]) {
          result += (cnt * y);
        } else {
          cout << "-1" << endl;
          return 0;
        }
      } else {
        if (maxi < prev || maxi < a[i]) {
          int var = cnt / k;
          result += min(var * x, var * k * y);
          result += y * (cnt % k);
        } else {
          result += x;
          cnt = cnt - k;
          int var = cnt / k;
          result += min(var * x, var * k * y);
          result += y * (cnt % k);
        }
      }
      maxi = 0;
      prev = a[i];
      cnt = 0;
    } else {
      cnt++;
      maxi = max(maxi, a[i]);
    }
  }
  if (j < m) {
    cout << "-1" << endl;
    return 0;
  }
  cout << result << endl;
}

```