#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pp = pair<int, int>;
const ll mod = 1000000007;
template <class T>
void print_arr(T* arr, int nn) {
  for (int i = 0; i < nn; ++i) {
    cout << arr[i] << ' ';
  }
  cout << endl;
}
unordered_map<char, char> beat = {{'R', 'P'}, {'S', 'R'}, {'P', 'S'}};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  ll x, k, y;
  cin >> x >> k >> y;
  int a[n], b[m];
  bool cheat = k * y < x;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < m; ++i) {
    cin >> b[i];
  }
  ll res = 0;
  int l = -1, r = 0;
  for (int i = 0; i < m; ++i) {
    int mx = 0;
    while (r < n && a[r] != b[i]) {
      mx = max(mx, a[r]);
      ++r;
    }
    if (r == n) {
      cout << -1 << endl;
      return 0;
    }
    int d = r - l - 1;
    int mxx = a[r];
    if (l >= 0) {
      mxx = max(a[l], a[r]);
    }
    if (d >= k) {
      if (cheat) {
        if (mxx > mx) {
          res += d * y;
        } else {
          res += x + (d - k) * y;
        }
      } else {
        res += x * (d / k) + y * (d % k);
      }
    } else {
      if (mxx < mx) {
        cout << -1 << endl;
        return 0;
      }
      res += y * d;
    }
    l = r;
    ++r;
  }
  int mx = 0;
  while (r != n) {
    mx = max(a[r], mx);
    ++r;
  }
  int d = r - l - 1;
  int mxx = a[l];
  if (d >= k) {
    if (cheat) {
      if (mxx > mx) {
        res += d * y;
      } else {
        res += x + (d - k) * y;
      }
    } else {
      res += x * (d / k) + y * (d % k);
    }
  } else {
    if (mxx < mx) {
      cout << -1 << endl;
      return 0;
    }
    res += y * d;
  }
  cout << res << 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;
using ll = long long;
using pp = pair<int, int>;
const ll mod = 1000000007;
template <class T>
void print_arr(T* arr, int nn) {
  for (int i = 0; i < nn; ++i) {
    cout << arr[i] << ' ';
  }
  cout << endl;
}
unordered_map<char, char> beat = {{'R', 'P'}, {'S', 'R'}, {'P', 'S'}};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  ll x, k, y;
  cin >> x >> k >> y;
  int a[n], b[m];
  bool cheat = k * y < x;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < m; ++i) {
    cin >> b[i];
  }
  ll res = 0;
  int l = -1, r = 0;
  for (int i = 0; i < m; ++i) {
    int mx = 0;
    while (r < n && a[r] != b[i]) {
      mx = max(mx, a[r]);
      ++r;
    }
    if (r == n) {
      cout << -1 << endl;
      return 0;
    }
    int d = r - l - 1;
    int mxx = a[r];
    if (l >= 0) {
      mxx = max(a[l], a[r]);
    }
    if (d >= k) {
      if (cheat) {
        if (mxx > mx) {
          res += d * y;
        } else {
          res += x + (d - k) * y;
        }
      } else {
        res += x * (d / k) + y * (d % k);
      }
    } else {
      if (mxx < mx) {
        cout << -1 << endl;
        return 0;
      }
      res += y * d;
    }
    l = r;
    ++r;
  }
  int mx = 0;
  while (r != n) {
    mx = max(a[r], mx);
    ++r;
  }
  int d = r - l - 1;
  int mxx = a[l];
  if (d >= k) {
    if (cheat) {
      if (mxx > mx) {
        res += d * y;
      } else {
        res += x + (d - k) * y;
      }
    } else {
      res += x * (d / k) + y * (d % k);
    }
  } else {
    if (mxx < mx) {
      cout << -1 << endl;
      return 0;
    }
    res += y * d;
  }
  cout << res << endl;
  return 0;
}

```