#include <bits/stdc++.h>
using namespace std;
int x, k, y;
long long passed(vector<int> &arr, int range) {
  vector<int> pos;
  int n = arr.size();
  for (int i = 0; i < n; i++) {
    if (arr[i] > range) {
      pos.push_back(i);
    }
  }
  if (pos.size() and k > n) return -1;
  if (k * y <= x) {
    if (pos.size() == 0)
      return (long long)y * (long long)n;
    else {
      return (long long)(n - k) * (long long)y + (long long)x;
    }
  }
  {
    long long bz = n / k;
    long long r = n - bz * k;
    return bz * (long long)x + r * y;
  }
}
int main() {
  int n, m;
  cin >> n >> m;
  cin >> x >> k >> y;
  vector<int> a(n), b(m);
  map<int, bool> bm;
  for (int &i : a) cin >> i;
  for (int &i : b) {
    cin >> i;
    bm[i] = 1;
  }
  vector<int> c;
  for (int i : a) {
    if (bm[i]) c.push_back(i);
  }
  if (c.size() != b.size()) {
    cout << -1 << endl;
    return 0;
  }
  for (int i = 0; i < b.size(); i++) {
    if (b[i] != c[i]) {
      cout << -1 << endl;
      return 0;
    }
  }
  a.push_back(INT_MIN);
  b.push_back(INT_MIN);
  bm[INT_MIN] = 1;
  int idx = 0;
  int last = INT_MIN;
  long long ans = 0;
  while (idx < n) {
    vector<int> tmp;
    for (; idx < n; idx++) {
      if (bm[a[idx]]) {
        break;
      } else
        tmp.push_back(a[idx]);
    }
    if (tmp.size() > 0) {
      long long ret = passed(tmp, max(last, a[idx]));
      if (ret == -1) {
        cout << -1 << endl;
        return 0;
      } else {
        ans += ret;
      }
    }
    last = a[idx++];
  }
  cout << ans << endl;
}

### Prompt

Please create a solution in CPP 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 x, k, y;
long long passed(vector<int> &arr, int range) {
  vector<int> pos;
  int n = arr.size();
  for (int i = 0; i < n; i++) {
    if (arr[i] > range) {
      pos.push_back(i);
    }
  }
  if (pos.size() and k > n) return -1;
  if (k * y <= x) {
    if (pos.size() == 0)
      return (long long)y * (long long)n;
    else {
      return (long long)(n - k) * (long long)y + (long long)x;
    }
  }
  {
    long long bz = n / k;
    long long r = n - bz * k;
    return bz * (long long)x + r * y;
  }
}
int main() {
  int n, m;
  cin >> n >> m;
  cin >> x >> k >> y;
  vector<int> a(n), b(m);
  map<int, bool> bm;
  for (int &i : a) cin >> i;
  for (int &i : b) {
    cin >> i;
    bm[i] = 1;
  }
  vector<int> c;
  for (int i : a) {
    if (bm[i]) c.push_back(i);
  }
  if (c.size() != b.size()) {
    cout << -1 << endl;
    return 0;
  }
  for (int i = 0; i < b.size(); i++) {
    if (b[i] != c[i]) {
      cout << -1 << endl;
      return 0;
    }
  }
  a.push_back(INT_MIN);
  b.push_back(INT_MIN);
  bm[INT_MIN] = 1;
  int idx = 0;
  int last = INT_MIN;
  long long ans = 0;
  while (idx < n) {
    vector<int> tmp;
    for (; idx < n; idx++) {
      if (bm[a[idx]]) {
        break;
      } else
        tmp.push_back(a[idx]);
    }
    if (tmp.size() > 0) {
      long long ret = passed(tmp, max(last, a[idx]));
      if (ret == -1) {
        cout << -1 << endl;
        return 0;
      } else {
        ans += ret;
      }
    }
    last = a[idx++];
  }
  cout << ans << endl;
}

```