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

```