#include <bits/stdc++.h>
using namespace std;
long long int check(vector<long long int> v, long long int ma, long long int x,
                    long long int k, long long int y) {
  long long int n = v.size(), cnt = 0;
  if (!n) return 0;
  sort(v.begin(), v.end());
  if (v[n - 1] > ma && n < k) return -1;
  bool val = (v[n - 1] < ma ? true : false);
  cnt += (n % k) * y;
  n -= n % k;
  if (!n) return cnt;
  if (y * k >= x)
    cnt += (n / k * x);
  else if (val)
    cnt += n * y;
  else
    cnt += (n - k) * y + x;
  return cnt;
}
void solve() {
  long long int n, m, x, k, y;
  cin >> n >> m >> x >> k >> y;
  long long int a[n], b[m];
  for (long long int i = 0; i < n; i++) cin >> a[i];
  for (long long int i = 0; i < m; i++) cin >> b[i];
  long long int prev = -1;
  vector<long long int> v;
  long long int cnt = 0, ki = 0;
  for (long long int i = 0; i < n; i++) {
    if (a[i] != b[ki])
      v.push_back(a[i]);
    else {
      long long int ma = max(prev, a[i]), xi = check(v, ma, x, k, y);
      if (xi == -1) {
        cout << -1 << endl;
        return;
      } else
        cnt += xi;
      v.clear();
      prev = a[i], ++ki;
    }
  }
  if (ki < m)
    cout << -1 << endl;
  else {
    long long int xi = check(v, prev, x, k, y);
    if (xi == -1) {
      cout << -1 << endl;
      return;
    } else
      cnt += xi;
    cout << cnt << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  solve();
  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 check(vector<long long int> v, long long int ma, long long int x,
                    long long int k, long long int y) {
  long long int n = v.size(), cnt = 0;
  if (!n) return 0;
  sort(v.begin(), v.end());
  if (v[n - 1] > ma && n < k) return -1;
  bool val = (v[n - 1] < ma ? true : false);
  cnt += (n % k) * y;
  n -= n % k;
  if (!n) return cnt;
  if (y * k >= x)
    cnt += (n / k * x);
  else if (val)
    cnt += n * y;
  else
    cnt += (n - k) * y + x;
  return cnt;
}
void solve() {
  long long int n, m, x, k, y;
  cin >> n >> m >> x >> k >> y;
  long long int a[n], b[m];
  for (long long int i = 0; i < n; i++) cin >> a[i];
  for (long long int i = 0; i < m; i++) cin >> b[i];
  long long int prev = -1;
  vector<long long int> v;
  long long int cnt = 0, ki = 0;
  for (long long int i = 0; i < n; i++) {
    if (a[i] != b[ki])
      v.push_back(a[i]);
    else {
      long long int ma = max(prev, a[i]), xi = check(v, ma, x, k, y);
      if (xi == -1) {
        cout << -1 << endl;
        return;
      } else
        cnt += xi;
      v.clear();
      prev = a[i], ++ki;
    }
  }
  if (ki < m)
    cout << -1 << endl;
  else {
    long long int xi = check(v, prev, x, k, y);
    if (xi == -1) {
      cout << -1 << endl;
      return;
    } else
      cnt += xi;
    cout << cnt << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}

```