#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int n, m;
  cin >> n >> m;
  long long int x, k, y;
  cin >> x >> k >> y;
  long long int a[n], b[m];
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < m; i++) cin >> b[i];
  int in = -1;
  long long int ans = 0, ct = 0, mx = 0;
  for (int i = 0; i < m; i++) {
    long long int ck = 0;
    ct = 0;
    mx = 0;
    for (int j = in + 1; j < n; j++) {
      in = j;
      if (b[i] == a[j]) {
        ck = 1;
        if (ct == 0) {
          break;
        }
        if (k * y < x) {
          if ((i != 0 and mx < b[i - 1]) or mx < a[j]) {
            ans += (y * ct);
          } else if (ct >= k) {
            ans += x;
            ans += (ct - k) * y;
          } else {
            cout << -1 << endl;
            return 0;
          }
        } else {
          if (ct >= k) {
            ans += ((ct / k) * x);
            ans += (ct % k) * y;
          } else if ((i != 0 and mx < b[i - 1]) or mx < a[j]) {
            ans += (y * ct);
          } else {
            cout << -1 << endl;
            return 0;
          }
        }
        break;
      } else {
        mx = max(mx, a[j]);
        ++ct;
      }
    }
    if (ck == 0) {
      cout << -1 << endl;
      return 0;
    }
  }
  ct = 0, mx = 0;
  for (int i = in + 1; i < n; i++) {
    ++ct;
    mx = max(mx, a[i]);
  }
  if (ct > 0) {
    if (k * y < x) {
      if (mx < b[m - 1]) {
        ans += (y * ct);
      } else if (ct >= k) {
        ans += x;
        ans += ((ct - k) * y);
      } else {
        cout << -1 << endl;
        return 0;
      }
    } else {
      if (ct >= k) {
        ans += ((ct / k) * x);
        ans += ((ct % k) * y);
      } else if (mx < b[m - 1]) {
        ans += (y * ct);
      } else {
        cout << -1 << endl;
        return 0;
      }
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Develop a solution in cpp 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;
int main() {
  long long int n, m;
  cin >> n >> m;
  long long int x, k, y;
  cin >> x >> k >> y;
  long long int a[n], b[m];
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < m; i++) cin >> b[i];
  int in = -1;
  long long int ans = 0, ct = 0, mx = 0;
  for (int i = 0; i < m; i++) {
    long long int ck = 0;
    ct = 0;
    mx = 0;
    for (int j = in + 1; j < n; j++) {
      in = j;
      if (b[i] == a[j]) {
        ck = 1;
        if (ct == 0) {
          break;
        }
        if (k * y < x) {
          if ((i != 0 and mx < b[i - 1]) or mx < a[j]) {
            ans += (y * ct);
          } else if (ct >= k) {
            ans += x;
            ans += (ct - k) * y;
          } else {
            cout << -1 << endl;
            return 0;
          }
        } else {
          if (ct >= k) {
            ans += ((ct / k) * x);
            ans += (ct % k) * y;
          } else if ((i != 0 and mx < b[i - 1]) or mx < a[j]) {
            ans += (y * ct);
          } else {
            cout << -1 << endl;
            return 0;
          }
        }
        break;
      } else {
        mx = max(mx, a[j]);
        ++ct;
      }
    }
    if (ck == 0) {
      cout << -1 << endl;
      return 0;
    }
  }
  ct = 0, mx = 0;
  for (int i = in + 1; i < n; i++) {
    ++ct;
    mx = max(mx, a[i]);
  }
  if (ct > 0) {
    if (k * y < x) {
      if (mx < b[m - 1]) {
        ans += (y * ct);
      } else if (ct >= k) {
        ans += x;
        ans += ((ct - k) * y);
      } else {
        cout << -1 << endl;
        return 0;
      }
    } else {
      if (ct >= k) {
        ans += ((ct / k) * x);
        ans += ((ct % k) * y);
      } else if (mx < b[m - 1]) {
        ans += (y * ct);
      } else {
        cout << -1 << endl;
        return 0;
      }
    }
  }
  cout << ans << endl;
  return 0;
}

```