#include <bits/stdc++.h>
using namespace std;
const int64_t M1 = 998244353;
const int64_t M2 = 1000000007;
void solve() {
  int64_t n, m, x, k, y;
  cin >> n >> m >> x >> k >> y;
  vector<int64_t> a(n + 2), b(m + 1);
  for (int64_t i = 1; i <= n; i++) cin >> a[i];
  for (int64_t i = 0; i < m; i++) cin >> b[i];
  a[0] = a[n + 1] = b[m] = -1;
  int64_t l = 0, mx = -1, ans = 0, pre = 0;
  for (int64_t i = 1; i <= n + 1; i++) {
    if (a[i] == b[l]) {
      if (i - pre - 1 < k) {
        if (mx > max(a[pre], a[i])) {
          cout << "-1\n";
          return;
        } else
          ans += (i - pre - 1) * y;
      } else {
        int64_t d = (i - pre - 1);
        ans += (d % k) * y;
        d -= d % k;
        if (mx > max(a[pre], a[i])) {
          ans += x;
          d -= k;
        }
        ans += min(d * y, (d / k) * x);
      }
      pre = i;
      l++;
      mx = -1;
    } else {
      mx = max(mx, a[i]);
    }
  }
  if (l <= m) ans = -1;
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  solve();
}

### Prompt

Create a solution in Cpp for 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;
const int64_t M1 = 998244353;
const int64_t M2 = 1000000007;
void solve() {
  int64_t n, m, x, k, y;
  cin >> n >> m >> x >> k >> y;
  vector<int64_t> a(n + 2), b(m + 1);
  for (int64_t i = 1; i <= n; i++) cin >> a[i];
  for (int64_t i = 0; i < m; i++) cin >> b[i];
  a[0] = a[n + 1] = b[m] = -1;
  int64_t l = 0, mx = -1, ans = 0, pre = 0;
  for (int64_t i = 1; i <= n + 1; i++) {
    if (a[i] == b[l]) {
      if (i - pre - 1 < k) {
        if (mx > max(a[pre], a[i])) {
          cout << "-1\n";
          return;
        } else
          ans += (i - pre - 1) * y;
      } else {
        int64_t d = (i - pre - 1);
        ans += (d % k) * y;
        d -= d % k;
        if (mx > max(a[pre], a[i])) {
          ans += x;
          d -= k;
        }
        ans += min(d * y, (d / k) * x);
      }
      pre = i;
      l++;
      mx = -1;
    } else {
      mx = max(mx, a[i]);
    }
  }
  if (l <= m) ans = -1;
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  solve();
}

```