#include <bits/stdc++.h>
using namespace std;
long long int M = 1000000007;
int main() {
  long long int n, m;
  cin >> n >> m;
  long long int x, k, y;
  cin >> x >> k >> y;
  long long int arr[n + 2];
  for (long long int i = 1; i <= n; i++) cin >> arr[i];
  arr[0] = arr[n + 1] = 0;
  long long int tr[m + 2];
  tr[0] = 0;
  tr[m + 1] = 0;
  for (long long int i = 1; i <= m; i++) cin >> tr[i];
  long long int ind = 1;
  long long int ans = 0;
  long long int cc = 0;
  for (long long int i = 1; i <= m + 1; i++) {
    if (ind > n + 1) {
      cc = 1;
      break;
    }
    long long int ch = max(tr[i - 1], tr[i]);
    long long int sm = 0, gr = 0;
    while (ind <= n + 1) {
      if (arr[ind] == tr[i]) {
        ind++;
        break;
      }
      if (arr[ind] < ch)
        sm++;
      else
        gr++;
      ind++;
    }
    long long int range = gr + sm;
    if (gr == 0) {
      if (x <= y * k) {
        ans += (range % k) * y + (range / k) * x;
      } else {
        ans += range * y;
      }
    } else {
      if (range < k) {
        cc = 1;
        break;
      } else {
        if (x <= y * k) {
          ans += (range % k) * y + (range / k) * x;
        } else {
          ans += (range - k) * y + x;
        }
      }
    }
  }
  if (cc)
    cout << -1 << endl;
  else
    cout << ans << endl;
  return 0;
}

### 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;
long long int M = 1000000007;
int main() {
  long long int n, m;
  cin >> n >> m;
  long long int x, k, y;
  cin >> x >> k >> y;
  long long int arr[n + 2];
  for (long long int i = 1; i <= n; i++) cin >> arr[i];
  arr[0] = arr[n + 1] = 0;
  long long int tr[m + 2];
  tr[0] = 0;
  tr[m + 1] = 0;
  for (long long int i = 1; i <= m; i++) cin >> tr[i];
  long long int ind = 1;
  long long int ans = 0;
  long long int cc = 0;
  for (long long int i = 1; i <= m + 1; i++) {
    if (ind > n + 1) {
      cc = 1;
      break;
    }
    long long int ch = max(tr[i - 1], tr[i]);
    long long int sm = 0, gr = 0;
    while (ind <= n + 1) {
      if (arr[ind] == tr[i]) {
        ind++;
        break;
      }
      if (arr[ind] < ch)
        sm++;
      else
        gr++;
      ind++;
    }
    long long int range = gr + sm;
    if (gr == 0) {
      if (x <= y * k) {
        ans += (range % k) * y + (range / k) * x;
      } else {
        ans += range * y;
      }
    } else {
      if (range < k) {
        cc = 1;
        break;
      } else {
        if (x <= y * k) {
          ans += (range % k) * y + (range / k) * x;
        } else {
          ans += (range - k) * y + x;
        }
      }
    }
  }
  if (cc)
    cout << -1 << endl;
  else
    cout << ans << endl;
  return 0;
}

```