#include <bits/stdc++.h>
using namespace std;
void fastio(void);
long long a[200005], b[200005], idx[200005];
int main() {
  fastio();
  long long n, m, x, y, k, la, lb;
  cin >> n >> m >> x >> k >> y;
  idx[0] = 0, idx[n + 1] = n + 1;
  for (la = 1; la <= n; la++) cin >> a[la], idx[a[la]] = la;
  for (la = 1; la <= m; la++) cin >> b[la];
  b[0] = 0, b[m + 1] = 0;
  long long ans = 0;
  bool mk = true;
  for (la = 1; la <= m + 1; la++) {
    if (idx[b[la]] < idx[b[la - 1]]) {
      mk = false;
      break;
    }
    long long mx = 0;
    for (lb = idx[b[la - 1]] + 1; lb < idx[b[la]]; lb++) mx = max(a[lb], mx);
    long long rem = idx[b[la]] - idx[b[la - 1]] - 1;
    if ((mx > b[la] && mx > b[la - 1]) && rem < k) {
      mk = false;
      break;
    }
    if (mx >= b[la] && mx >= b[la - 1]) {
      ans += x;
      rem -= k;
    }
    ans += (rem % k) * y;
    rem -= (rem % k);
    ans += min((rem / k) * x, rem * y);
    if (la == 1) idx[0] = n + 1;
  }
  if (!mk) ans = -1;
  cout << ans << endl;
  return 0;
}
void fastio(void) {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}

### Prompt

Your challenge is to write 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;
void fastio(void);
long long a[200005], b[200005], idx[200005];
int main() {
  fastio();
  long long n, m, x, y, k, la, lb;
  cin >> n >> m >> x >> k >> y;
  idx[0] = 0, idx[n + 1] = n + 1;
  for (la = 1; la <= n; la++) cin >> a[la], idx[a[la]] = la;
  for (la = 1; la <= m; la++) cin >> b[la];
  b[0] = 0, b[m + 1] = 0;
  long long ans = 0;
  bool mk = true;
  for (la = 1; la <= m + 1; la++) {
    if (idx[b[la]] < idx[b[la - 1]]) {
      mk = false;
      break;
    }
    long long mx = 0;
    for (lb = idx[b[la - 1]] + 1; lb < idx[b[la]]; lb++) mx = max(a[lb], mx);
    long long rem = idx[b[la]] - idx[b[la - 1]] - 1;
    if ((mx > b[la] && mx > b[la - 1]) && rem < k) {
      mk = false;
      break;
    }
    if (mx >= b[la] && mx >= b[la - 1]) {
      ans += x;
      rem -= k;
    }
    ans += (rem % k) * y;
    rem -= (rem % k);
    ans += min((rem / k) * x, rem * y);
    if (la == 1) idx[0] = n + 1;
  }
  if (!mk) ans = -1;
  cout << ans << endl;
  return 0;
}
void fastio(void) {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}

```