#include <bits/stdc++.h>
using namespace std;
int n, m, a[200001], b[200001];
long long x, k, y;
void read() {
  cin >> n >> m >> x >> k >> y;
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < m; ++i) cin >> b[i];
}
bool update(int l, int r, long long &res) {
  if (l > r) return true;
  bool ok = false;
  int maxx = *max_element(a + l, a + r + 1);
  int lr = r - l + 1;
  if ((l - 1 >= 0 && a[l - 1] > maxx) || (r + 1 < n && a[r + 1] > maxx))
    ok = true;
  if (lr < k && !ok) return false;
  int need = lr % k;
  res += need * y;
  lr -= need;
  if (y * k >= x)
    res += lr / k * x;
  else if (ok)
    res += lr * y;
  else
    res += (lr - k) * y + x;
  return true;
}
int main() {
  read();
  long long res = 0;
  int posA = 0, posB = 0, last = -1;
  while (posB < m) {
    while (posA < n && a[posA] != b[posB]) ++posA;
    if (posA == n) {
      cout << -1;
      return 0;
    }
    if (!update(last + 1, posA - 1, res)) {
      cout << -1;
      return 0;
    }
    last = posA;
    ++posB;
  }
  if (!update(last + 1, n - 1, res)) {
    cout << -1;
    return 0;
  }
  cout << res;
  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 n, m, a[200001], b[200001];
long long x, k, y;
void read() {
  cin >> n >> m >> x >> k >> y;
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < m; ++i) cin >> b[i];
}
bool update(int l, int r, long long &res) {
  if (l > r) return true;
  bool ok = false;
  int maxx = *max_element(a + l, a + r + 1);
  int lr = r - l + 1;
  if ((l - 1 >= 0 && a[l - 1] > maxx) || (r + 1 < n && a[r + 1] > maxx))
    ok = true;
  if (lr < k && !ok) return false;
  int need = lr % k;
  res += need * y;
  lr -= need;
  if (y * k >= x)
    res += lr / k * x;
  else if (ok)
    res += lr * y;
  else
    res += (lr - k) * y + x;
  return true;
}
int main() {
  read();
  long long res = 0;
  int posA = 0, posB = 0, last = -1;
  while (posB < m) {
    while (posA < n && a[posA] != b[posB]) ++posA;
    if (posA == n) {
      cout << -1;
      return 0;
    }
    if (!update(last + 1, posA - 1, res)) {
      cout << -1;
      return 0;
    }
    last = posA;
    ++posB;
  }
  if (!update(last + 1, n - 1, res)) {
    cout << -1;
    return 0;
  }
  cout << res;
  return 0;
}

```