#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, m, x, y, k, ans = 0;
  cin >> n >> m;
  cin >> x >> k >> y;
  vector<long long> a(n + 2, -1), b(m + 2, -1), p, g;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int j = 1; j <= m; j++) cin >> b[j];
  long long l = 0, last = 0;
  for (int i = 1; i <= m + 1; i++) {
    last = l;
    while (a[l + 1] != b[i] && l + 1 <= n + 1) l++;
    l++;
    if (l == n + 2) {
      cout << -1;
      return 0;
    }
    long long ind1 = 0;
    for (int j = last; j <= l; j++) {
      ind1 = max(a[j], ind1);
    }
    long long kol = l - last - 1;
    if (ind1 == a[last] || ind1 == a[l]) {
      while (kol % k != 0) {
        kol--;
        ans += y;
      }
      ans += kol / k * min(x, y * k);
      continue;
    }
    if (kol == 0) continue;
    if (kol < k) {
      cout << -1;
      return 0;
    }
    ans += x;
    kol -= k;
    while (kol % k != 0) {
      kol--;
      ans += y;
    }
    ans += kol / k * min(x, y * k);
  }
  cout << ans << "\n";
}

### Prompt

Please provide a cpp coded solution 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 main() {
  long long n, m, x, y, k, ans = 0;
  cin >> n >> m;
  cin >> x >> k >> y;
  vector<long long> a(n + 2, -1), b(m + 2, -1), p, g;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int j = 1; j <= m; j++) cin >> b[j];
  long long l = 0, last = 0;
  for (int i = 1; i <= m + 1; i++) {
    last = l;
    while (a[l + 1] != b[i] && l + 1 <= n + 1) l++;
    l++;
    if (l == n + 2) {
      cout << -1;
      return 0;
    }
    long long ind1 = 0;
    for (int j = last; j <= l; j++) {
      ind1 = max(a[j], ind1);
    }
    long long kol = l - last - 1;
    if (ind1 == a[last] || ind1 == a[l]) {
      while (kol % k != 0) {
        kol--;
        ans += y;
      }
      ans += kol / k * min(x, y * k);
      continue;
    }
    if (kol == 0) continue;
    if (kol < k) {
      cout << -1;
      return 0;
    }
    ans += x;
    kol -= k;
    while (kol % k != 0) {
      kol--;
      ans += y;
    }
    ans += kol / k * min(x, y * k);
  }
  cout << ans << "\n";
}

```