#include <bits/stdc++.h>
using namespace std;
long long n, m, x, y, k, a[200005], b[200005], w[200005], nr;
int main() {
  long long i, j, ans = 0;
  cin >> n >> m;
  cin >> x >> k >> y;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= m; i++) cin >> b[i];
  i = 1;
  j = 1;
  ans = 0;
  long long ok = 1;
  long long nrelem = 0, retine, maxi1 = 0, maxi2 = 0, poz = 0;
  a[n + 1] = -1;
  b[m + 1] = -1;
  while (i <= n && j <= m && ok == 1) {
    if (a[i] != b[j]) {
      if (a[i] > maxi2) maxi2 = a[i];
      i++;
    } else {
      maxi1 = max(maxi1, a[i]);
      long long dif = i - poz - 1;
      if (maxi1 >= maxi2)
        ans += min((x * (dif / k) + (dif % k) * y), y * dif);
      else {
        if (dif < k)
          j = -1;
        else
          ans += min(x * (dif / k) + (dif % k) * y, x + (dif - k) * y);
      }
      poz = i;
      maxi1 = a[i];
      maxi2 = 0;
      j++;
      i++;
    }
  }
  if (j < m)
    cout << -1;
  else {
    maxi2 = 0;
    retine = i;
    ok = 1;
    while (i <= n) {
      if (a[i] > maxi2) maxi2 = a[i];
      i++;
    }
    long long dif = n - poz;
    if (maxi1 >= maxi2)
      ans += min((x * (dif / k) + (dif % k) * y), y * dif);
    else {
      if (dif < k)
        ok = 0;
      else
        ans += min(x * (dif / k) + (dif % k) * y, x + (dif - k) * y);
    }
    if (ok == 1)
      cout << ans;
    else
      cout << -1;
  }
  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 n, m, x, y, k, a[200005], b[200005], w[200005], nr;
int main() {
  long long i, j, ans = 0;
  cin >> n >> m;
  cin >> x >> k >> y;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= m; i++) cin >> b[i];
  i = 1;
  j = 1;
  ans = 0;
  long long ok = 1;
  long long nrelem = 0, retine, maxi1 = 0, maxi2 = 0, poz = 0;
  a[n + 1] = -1;
  b[m + 1] = -1;
  while (i <= n && j <= m && ok == 1) {
    if (a[i] != b[j]) {
      if (a[i] > maxi2) maxi2 = a[i];
      i++;
    } else {
      maxi1 = max(maxi1, a[i]);
      long long dif = i - poz - 1;
      if (maxi1 >= maxi2)
        ans += min((x * (dif / k) + (dif % k) * y), y * dif);
      else {
        if (dif < k)
          j = -1;
        else
          ans += min(x * (dif / k) + (dif % k) * y, x + (dif - k) * y);
      }
      poz = i;
      maxi1 = a[i];
      maxi2 = 0;
      j++;
      i++;
    }
  }
  if (j < m)
    cout << -1;
  else {
    maxi2 = 0;
    retine = i;
    ok = 1;
    while (i <= n) {
      if (a[i] > maxi2) maxi2 = a[i];
      i++;
    }
    long long dif = n - poz;
    if (maxi1 >= maxi2)
      ans += min((x * (dif / k) + (dif % k) * y), y * dif);
    else {
      if (dif < k)
        ok = 0;
      else
        ans += min(x * (dif / k) + (dif % k) * y, x + (dif - k) * y);
    }
    if (ok == 1)
      cout << ans;
    else
      cout << -1;
  }
  return 0;
}

```