#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n, m, k;
  cin >> n >> m;
  long long int x, y;
  cin >> x >> k >> y;
  long long int a[n], b[m];
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < m; i++) cin >> b[i];
  long long int ans = 0;
  int l = -1, r = 0;
  int i = 0, j = 0;
  while (j < m) {
    while (i < n && a[i] != b[j]) i++;
    if (i == n) {
      cout << -1;
      return;
    }
    r = i;
    if (l + 1 > r - 1) {
      j++;
      l = i;
      continue;
    }
    long long int mx = 0;
    for (int h = l + 1; h < r; h++) mx = max(mx, a[h]);
    int q = 0;
    if (l >= 0 && a[l] > mx) {
      q = 1;
    }
    if (r < n && a[r] > mx) {
      q = 1;
    }
    int d = r - l - 1;
    if (q == 0 && d < k) {
      cout << -1;
      return;
    }
    int p = d % k;
    ans += p * y;
    d -= p;
    if (x <= k * y)
      ans += (d / k) * x;
    else if (q == 0)
      ans += (d - k) * y + x;
    else {
      ans += d * y;
    }
    l = i;
    j++;
  }
  r = n;
  if (l + 1 > r - 1) {
    cout << ans;
    return;
  }
  long long int mx = 0;
  for (int h = l + 1; h < r; h++) mx = max(mx, a[h]);
  int q = 0;
  if (l >= 0 && a[l] > mx) {
    q = 1;
  }
  if (r < n && a[r] > mx) {
    q = 1;
  }
  int d = r - l - 1;
  if (q == 0 && d < k) {
    cout << -1;
    return;
  }
  int p = d % k;
  ans += p * y;
  d -= p;
  if (x <= k * y)
    ans += (d / k) * x;
  else if (q == 0)
    ans += (d - k) * y + x;
  else {
    ans += d * y;
  }
  cout << ans << endl;
}
int main() {
  int t;
  t = 1;
  while (t--) solve();
}

### Prompt

Please create a solution in CPP 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 solve() {
  int n, m, k;
  cin >> n >> m;
  long long int x, y;
  cin >> x >> k >> y;
  long long int a[n], b[m];
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < m; i++) cin >> b[i];
  long long int ans = 0;
  int l = -1, r = 0;
  int i = 0, j = 0;
  while (j < m) {
    while (i < n && a[i] != b[j]) i++;
    if (i == n) {
      cout << -1;
      return;
    }
    r = i;
    if (l + 1 > r - 1) {
      j++;
      l = i;
      continue;
    }
    long long int mx = 0;
    for (int h = l + 1; h < r; h++) mx = max(mx, a[h]);
    int q = 0;
    if (l >= 0 && a[l] > mx) {
      q = 1;
    }
    if (r < n && a[r] > mx) {
      q = 1;
    }
    int d = r - l - 1;
    if (q == 0 && d < k) {
      cout << -1;
      return;
    }
    int p = d % k;
    ans += p * y;
    d -= p;
    if (x <= k * y)
      ans += (d / k) * x;
    else if (q == 0)
      ans += (d - k) * y + x;
    else {
      ans += d * y;
    }
    l = i;
    j++;
  }
  r = n;
  if (l + 1 > r - 1) {
    cout << ans;
    return;
  }
  long long int mx = 0;
  for (int h = l + 1; h < r; h++) mx = max(mx, a[h]);
  int q = 0;
  if (l >= 0 && a[l] > mx) {
    q = 1;
  }
  if (r < n && a[r] > mx) {
    q = 1;
  }
  int d = r - l - 1;
  if (q == 0 && d < k) {
    cout << -1;
    return;
  }
  int p = d % k;
  ans += p * y;
  d -= p;
  if (x <= k * y)
    ans += (d / k) * x;
  else if (q == 0)
    ans += (d - k) * y + x;
  else {
    ans += d * y;
  }
  cout << ans << endl;
}
int main() {
  int t;
  t = 1;
  while (t--) solve();
}

```