#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 7;
int64_t a[N], b[N];
int main() {
  int64_t n, m, x, k, y;
  cin >> n >> m >> x >> k >> y;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= m; i++) cin >> b[i];
  int64_t pos = 1, endpoint = 0, cnt = 0, maxi = 0, brk = 0, sum = 0;
  for (int i = 1; i <= n + 1; i++) {
    if (a[i] == b[pos]) {
      pos++;
      endpoint = max(endpoint, a[i]);
      if (cnt < k and endpoint < maxi) brk = -1;
      int64_t y_num = cnt % k;
      sum += y_num * y;
      cnt -= y_num;
      if (maxi > endpoint) {
        sum += x;
        cnt -= k;
      }
      sum += min(cnt * y, cnt / k * x);
      cnt = 0;
      maxi = 0;
      endpoint = a[i];
    } else {
      cnt++;
      maxi = max(maxi, a[i]);
    }
  }
  if (brk == -1 || pos != m + 2) {
    cout << -1;
  } else {
    cout << sum;
  }
}

### 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 N = 2e5 + 7;
int64_t a[N], b[N];
int main() {
  int64_t n, m, x, k, y;
  cin >> n >> m >> x >> k >> y;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= m; i++) cin >> b[i];
  int64_t pos = 1, endpoint = 0, cnt = 0, maxi = 0, brk = 0, sum = 0;
  for (int i = 1; i <= n + 1; i++) {
    if (a[i] == b[pos]) {
      pos++;
      endpoint = max(endpoint, a[i]);
      if (cnt < k and endpoint < maxi) brk = -1;
      int64_t y_num = cnt % k;
      sum += y_num * y;
      cnt -= y_num;
      if (maxi > endpoint) {
        sum += x;
        cnt -= k;
      }
      sum += min(cnt * y, cnt / k * x);
      cnt = 0;
      maxi = 0;
      endpoint = a[i];
    } else {
      cnt++;
      maxi = max(maxi, a[i]);
    }
  }
  if (brk == -1 || pos != m + 2) {
    cout << -1;
  } else {
    cout << sum;
  }
}

```