#include <bits/stdc++.h>
using namespace std;
const int MAX_INT = 2147483647;
const long long MAX_LL = 9223372036854775807;
int main() {
  int n, m;
  cin >> n >> m;
  vector<long long> a(n), b(m);
  long long x, k, y;
  cin >> x >> k >> y;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < m; i++) cin >> b[i];
  a.push_back(-1);
  b.push_back(-1);
  int index = 0;
  long long ans = 0;
  int last = -1;
  long long maximum = -1;
  for (int i = 0; i <= n; i++) {
    if (a[i] == b[index]) {
      int between = i - last - 1;
      if (between >= k) {
        long long toAdd = (between / k) * x + (between % k) * y;
        if (maximum < b[index] || (index > 0 && maximum < b[index - 1])) {
          toAdd = min(toAdd, between * y);
        } else {
          toAdd = min(toAdd, x + (between - k) * y);
        }
        ans += toAdd;
      } else if (between != 0) {
        if (maximum < b[index] || (index > 0 && maximum < b[index - 1])) {
          ans += between * y;
        } else {
          ans = -1;
          break;
        }
      }
      ++index;
      maximum = -1;
      last = i;
    } else {
      maximum = max(maximum, a[i]);
    }
  }
  if (index != m + 1) ans = -1;
  cout << ans << endl;
  return 0;
}

### 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>
using namespace std;
const int MAX_INT = 2147483647;
const long long MAX_LL = 9223372036854775807;
int main() {
  int n, m;
  cin >> n >> m;
  vector<long long> a(n), b(m);
  long long x, k, y;
  cin >> x >> k >> y;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < m; i++) cin >> b[i];
  a.push_back(-1);
  b.push_back(-1);
  int index = 0;
  long long ans = 0;
  int last = -1;
  long long maximum = -1;
  for (int i = 0; i <= n; i++) {
    if (a[i] == b[index]) {
      int between = i - last - 1;
      if (between >= k) {
        long long toAdd = (between / k) * x + (between % k) * y;
        if (maximum < b[index] || (index > 0 && maximum < b[index - 1])) {
          toAdd = min(toAdd, between * y);
        } else {
          toAdd = min(toAdd, x + (between - k) * y);
        }
        ans += toAdd;
      } else if (between != 0) {
        if (maximum < b[index] || (index > 0 && maximum < b[index - 1])) {
          ans += between * y;
        } else {
          ans = -1;
          break;
        }
      }
      ++index;
      maximum = -1;
      last = i;
    } else {
      maximum = max(maximum, a[i]);
    }
  }
  if (index != m + 1) ans = -1;
  cout << ans << endl;
  return 0;
}

```