#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const int N = 2e5 + 5;
int n, m, a[N], b[N];
int x, k, y;
long long ans;
bool get(int l, int r) {
  if (l > r) return true;
  int cnt = r - l + 1, mx = 0, tmp = l;
  while (tmp <= r) mx = max(mx, a[tmp]), tmp++;
  bool yes = (l - 1 >= 0 && a[l - 1] > mx);
  yes |= (r + 1 < n && a[r + 1] > mx);
  if (cnt < k && !yes) return false;
  ans += 1ll * (cnt % k) * y;
  cnt -= cnt % k;
  if (y * k >= x)
    ans += 1ll * (cnt / k) * x;
  else if (yes)
    ans += 1ll * cnt * y;
  else
    ans += 1ll * (cnt - k) * y + x;
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  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];
  int j = 0, lst = 0;
  for (int i = 0; i < m; i++) {
    while (j < n && a[j] != b[i]) j++;
    if (j == n || !get(lst, j - 1)) return cout << -1, 0;
    lst = j + 1;
  }
  if (!get(lst, n - 1)) return cout << -1, 0;
  cout << ans;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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>
#pragma GCC optimize("O3")
using namespace std;
const int N = 2e5 + 5;
int n, m, a[N], b[N];
int x, k, y;
long long ans;
bool get(int l, int r) {
  if (l > r) return true;
  int cnt = r - l + 1, mx = 0, tmp = l;
  while (tmp <= r) mx = max(mx, a[tmp]), tmp++;
  bool yes = (l - 1 >= 0 && a[l - 1] > mx);
  yes |= (r + 1 < n && a[r + 1] > mx);
  if (cnt < k && !yes) return false;
  ans += 1ll * (cnt % k) * y;
  cnt -= cnt % k;
  if (y * k >= x)
    ans += 1ll * (cnt / k) * x;
  else if (yes)
    ans += 1ll * cnt * y;
  else
    ans += 1ll * (cnt - k) * y + x;
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  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];
  int j = 0, lst = 0;
  for (int i = 0; i < m; i++) {
    while (j < n && a[j] != b[i]) j++;
    if (j == n || !get(lst, j - 1)) return cout << -1, 0;
    lst = j + 1;
  }
  if (!get(lst, n - 1)) return cout << -1, 0;
  cout << ans;
  return 0;
}

```