#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b) {
  if (b == 1) return a;
  if (b == 0) return 1;
  long long temp = power(a, b / 2);
  long long ans = temp * temp;
  if (b & 1) ans *= a;
  return ans;
}
long long power(long long a, long long b, long long p) {
  a = a % p;
  if (b == 1) return a;
  if (b == 0) return 1;
  long long temp = power(a, b / 2, p);
  long long ans = (temp * temp) % p;
  if (b & 1) ans = (a * ans) % p;
  return ans % p;
}
long long inv(long long a, long long m) { return power(a, m - 2, m); }
long long n, m;
long long x, k, y;
long long a[200001];
long long b[200001];
bool canDelete(long long st, long long en, long long &res) {
  if (st > en) return true;
  long long p = *max_element(a + st, a + en + 1);
  bool f = false;
  long long len = en - st + 1;
  if (st - 1 >= 1 and a[st - 1] > p) f = true;
  if (en + 1 <= n and a[en + 1] > p) f = true;
  if (len < k and !f) return false;
  long long rem = len % k;
  res += rem * y;
  len -= rem;
  if (y * k >= x)
    res += (len / k) * x;
  else if (f)
    res += len * y;
  else
    res += (len - k) * y + x;
  return true;
}
void solve() {
  cin >> n >> m;
  cin >> x >> k >> y;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 1; i <= m; i++) cin >> b[i];
  long long i = 1;
  long long prev = 0, now = 1, ans = 0;
  while (i <= m) {
    while (now <= n and a[now] != b[i]) now++;
    if (now == n + 1) {
      cout << -1 << '\n';
      return;
    }
    if (!canDelete(prev + 1, now - 1, ans)) {
      cout << -1 << '\n';
      return;
    }
    prev = now;
    i++;
  }
  if (!canDelete(prev + 1, n, ans)) {
    cout << -1 << '\n';
    return;
  }
  cout << ans << '\n';
}
signed main() {
  solve();
  return 0;
}

### Prompt

Please formulate 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;
long long power(long long a, long long b) {
  if (b == 1) return a;
  if (b == 0) return 1;
  long long temp = power(a, b / 2);
  long long ans = temp * temp;
  if (b & 1) ans *= a;
  return ans;
}
long long power(long long a, long long b, long long p) {
  a = a % p;
  if (b == 1) return a;
  if (b == 0) return 1;
  long long temp = power(a, b / 2, p);
  long long ans = (temp * temp) % p;
  if (b & 1) ans = (a * ans) % p;
  return ans % p;
}
long long inv(long long a, long long m) { return power(a, m - 2, m); }
long long n, m;
long long x, k, y;
long long a[200001];
long long b[200001];
bool canDelete(long long st, long long en, long long &res) {
  if (st > en) return true;
  long long p = *max_element(a + st, a + en + 1);
  bool f = false;
  long long len = en - st + 1;
  if (st - 1 >= 1 and a[st - 1] > p) f = true;
  if (en + 1 <= n and a[en + 1] > p) f = true;
  if (len < k and !f) return false;
  long long rem = len % k;
  res += rem * y;
  len -= rem;
  if (y * k >= x)
    res += (len / k) * x;
  else if (f)
    res += len * y;
  else
    res += (len - k) * y + x;
  return true;
}
void solve() {
  cin >> n >> m;
  cin >> x >> k >> y;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 1; i <= m; i++) cin >> b[i];
  long long i = 1;
  long long prev = 0, now = 1, ans = 0;
  while (i <= m) {
    while (now <= n and a[now] != b[i]) now++;
    if (now == n + 1) {
      cout << -1 << '\n';
      return;
    }
    if (!canDelete(prev + 1, now - 1, ans)) {
      cout << -1 << '\n';
      return;
    }
    prev = now;
    i++;
  }
  if (!canDelete(prev + 1, n, ans)) {
    cout << -1 << '\n';
    return;
  }
  cout << ans << '\n';
}
signed main() {
  solve();
  return 0;
}

```