#include <bits/stdc++.h>
using namespace std;
int n, m;
long long x, k, y;
int A[200003], B[200003];
bool bol = 1;
long long solve(vector<int> &vec, int mx) {
  int sz = vec.size();
  bool big = 0;
  for (int x : vec) big |= (x > mx);
  if (!vec.size()) return 0;
  if (vec.size() < k && big) {
    bol = 0;
    return 0;
  } else if (sz < k && !big)
    return y * sz;
  else if (sz >= k && big) {
    return y * (sz % k) + x + min((sz / k - 1) * x, (sz - sz % k - k) * y);
  } else
    return y * (sz % k) + min((sz / k) * x, (sz - sz % k) * y);
}
int main() {
  scanf("%d %d", &n, &m);
  scanf("%lld %lld %lld", &x, &k, &y);
  for (int i = 1; i <= n; i++) scanf("%d", &A[i]);
  for (int i = 1; i <= m; i++) scanf("%d", &B[i]);
  int mx = 0, j = 1;
  long long res = 0;
  vector<int> vec;
  for (int i = 1; i <= n; i++) {
    if (A[i] == B[j]) {
      mx = max(mx, B[j]);
      j++;
      res += solve(vec, mx);
      mx = A[i];
      vec.clear();
    } else {
      vec.push_back(A[i]);
    }
  }
  res += solve(vec, mx);
  if (j != m + 1) bol = 0;
  printf("%lld\n", (bol) ? res : -1);
}

### Prompt

Please formulate 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;
int n, m;
long long x, k, y;
int A[200003], B[200003];
bool bol = 1;
long long solve(vector<int> &vec, int mx) {
  int sz = vec.size();
  bool big = 0;
  for (int x : vec) big |= (x > mx);
  if (!vec.size()) return 0;
  if (vec.size() < k && big) {
    bol = 0;
    return 0;
  } else if (sz < k && !big)
    return y * sz;
  else if (sz >= k && big) {
    return y * (sz % k) + x + min((sz / k - 1) * x, (sz - sz % k - k) * y);
  } else
    return y * (sz % k) + min((sz / k) * x, (sz - sz % k) * y);
}
int main() {
  scanf("%d %d", &n, &m);
  scanf("%lld %lld %lld", &x, &k, &y);
  for (int i = 1; i <= n; i++) scanf("%d", &A[i]);
  for (int i = 1; i <= m; i++) scanf("%d", &B[i]);
  int mx = 0, j = 1;
  long long res = 0;
  vector<int> vec;
  for (int i = 1; i <= n; i++) {
    if (A[i] == B[j]) {
      mx = max(mx, B[j]);
      j++;
      res += solve(vec, mx);
      mx = A[i];
      vec.clear();
    } else {
      vec.push_back(A[i]);
    }
  }
  res += solve(vec, mx);
  if (j != m + 1) bol = 0;
  printf("%lld\n", (bol) ? res : -1);
}

```