#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, m;
  long long x, y, k;
  cin >> n >> m;
  cin >> x >> k >> y;
  vector<long long> arr;
  vector<long long> brr;
  arr.push_back(0);
  for (long long i = 0; i < n; i++) {
    long long num;
    cin >> num;
    arr.push_back(num);
  }
  arr.push_back(0);
  brr.push_back(0);
  for (long long i = 0; i < m; i++) {
    long long num;
    cin >> num;
    brr.push_back(num);
  }
  brr.push_back(0);
  vector<long long> v;
  long long ans = 0;
  for (long long i = 1, j = 1; i < arr.size(); i++) {
    if (arr[i] != brr[j]) {
      v.push_back(arr[i]);
    } else {
      if (v.size() == 0) {
        j++;
        continue;
      }
      sort(v.begin(), v.end());
      long long sz = v.size();
      long long mx = v[sz - 1];
      long long pash = max(brr[j], brr[j - 1]);
      if (pash < mx && sz < k) {
        cout << -1 << endl;
        return 0;
      } else if (k * y < x) {
        if (mx < pash) {
          ans += sz * y;
        } else
          ans += (sz - k) * y + x;
      } else {
        long long mod = sz % k;
        ans += (sz / k) * x;
        ans += mod * y;
      }
      j++;
      v.clear();
    }
    if (i + 1 == arr.size() && (j) != brr.size()) {
      cout << -1 << endl;
      return 0;
    }
  }
  cout << ans << endl;
}

### Prompt

Create a solution in cpp for 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;
int main() {
  long long n, m;
  long long x, y, k;
  cin >> n >> m;
  cin >> x >> k >> y;
  vector<long long> arr;
  vector<long long> brr;
  arr.push_back(0);
  for (long long i = 0; i < n; i++) {
    long long num;
    cin >> num;
    arr.push_back(num);
  }
  arr.push_back(0);
  brr.push_back(0);
  for (long long i = 0; i < m; i++) {
    long long num;
    cin >> num;
    brr.push_back(num);
  }
  brr.push_back(0);
  vector<long long> v;
  long long ans = 0;
  for (long long i = 1, j = 1; i < arr.size(); i++) {
    if (arr[i] != brr[j]) {
      v.push_back(arr[i]);
    } else {
      if (v.size() == 0) {
        j++;
        continue;
      }
      sort(v.begin(), v.end());
      long long sz = v.size();
      long long mx = v[sz - 1];
      long long pash = max(brr[j], brr[j - 1]);
      if (pash < mx && sz < k) {
        cout << -1 << endl;
        return 0;
      } else if (k * y < x) {
        if (mx < pash) {
          ans += sz * y;
        } else
          ans += (sz - k) * y + x;
      } else {
        long long mod = sz % k;
        ans += (sz / k) * x;
        ans += mod * y;
      }
      j++;
      v.clear();
    }
    if (i + 1 == arr.size() && (j) != brr.size()) {
      cout << -1 << endl;
      return 0;
    }
  }
  cout << ans << endl;
}

```