#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9 + 5, N = 5e4 + 5, mod = 1e9 + 7, M = 7e5 + 5;
int main() {
  long long n, m;
  cin >> n >> m;
  long long x, k, y;
  cin >> x >> k >> y;
  long long a[n + 5], b[m + 5];
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= m; i++) {
    cin >> b[i];
  }
  if (m > n) {
    cout << -1 << endl;
    return 0;
  }
  long long idx = 1, mx = 0, col = 0, last = 0, ans = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == b[idx]) {
      if (x < y * k) {
        if (col / k == 0) {
          if (mx > a[i] && mx > last) {
            cout << -1 << endl;
            return 0;
          }
          ans += (col * y);
        } else {
          long long pol = col / k, ost = (col % k);
          ans += (ost * y);
          ans += (pol * x);
        }
      }
      if (x == y * k) {
        if (col / k == 0) {
          if (mx > a[i] && mx > last) {
            cout << -1 << endl;
            return 0;
          }
          ans += (col * y);
        } else {
          long long pol = col / k, ost = (col % k);
          ans += (ost * y);
          ans += (pol * x);
        }
      }
      if (x > y * k) {
        if (col / k == 0) {
          if (mx > a[i] && mx > last) {
            cout << -1 << endl;
            return 0;
          }
          ans += (col * y);
        } else {
          if (mx > a[i] && mx > last) {
            long long ost = col - k;
            ans += x;
            ans += (ost * y);
          } else {
            ans += (col * y);
          }
        }
      }
      mx = 0;
      col = 0;
      last = a[i];
      idx++;
      continue;
    }
    mx = max(mx, a[i]);
    col++;
  }
  if (x < y * k) {
    if (col / k == 0) {
      if (mx > last) {
        cout << -1 << endl;
        return 0;
      }
      ans += (col * y);
    } else {
      long long pol = col / k, ost = (col % k);
      ans += (ost * y);
      ans += (pol * x);
    }
  }
  if (x == y * k) {
    if (col / k == 0) {
      if (mx > last) {
        cout << -1 << endl;
        return 0;
      }
      ans += (col * y);
    } else {
      long long pol = col / k, ost = (col % k);
      ans += (ost * y);
      ans += (pol * x);
    }
  }
  if (x > y * k) {
    if (col / k == 0) {
      if (mx > last) {
        cout << -1 << endl;
        return 0;
      }
      ans += (col * y);
    } else {
      if (mx > last) {
        long long ost = col - k;
        ans += x;
        ans += (ost * y);
      } else {
        ans += (col * y);
      }
    }
  }
  if (idx != m + 1) ans = -1;
  cout << ans << endl;
}

### Prompt

Please formulate 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;
const long long INF = 1e9 + 5, N = 5e4 + 5, mod = 1e9 + 7, M = 7e5 + 5;
int main() {
  long long n, m;
  cin >> n >> m;
  long long x, k, y;
  cin >> x >> k >> y;
  long long a[n + 5], b[m + 5];
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= m; i++) {
    cin >> b[i];
  }
  if (m > n) {
    cout << -1 << endl;
    return 0;
  }
  long long idx = 1, mx = 0, col = 0, last = 0, ans = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == b[idx]) {
      if (x < y * k) {
        if (col / k == 0) {
          if (mx > a[i] && mx > last) {
            cout << -1 << endl;
            return 0;
          }
          ans += (col * y);
        } else {
          long long pol = col / k, ost = (col % k);
          ans += (ost * y);
          ans += (pol * x);
        }
      }
      if (x == y * k) {
        if (col / k == 0) {
          if (mx > a[i] && mx > last) {
            cout << -1 << endl;
            return 0;
          }
          ans += (col * y);
        } else {
          long long pol = col / k, ost = (col % k);
          ans += (ost * y);
          ans += (pol * x);
        }
      }
      if (x > y * k) {
        if (col / k == 0) {
          if (mx > a[i] && mx > last) {
            cout << -1 << endl;
            return 0;
          }
          ans += (col * y);
        } else {
          if (mx > a[i] && mx > last) {
            long long ost = col - k;
            ans += x;
            ans += (ost * y);
          } else {
            ans += (col * y);
          }
        }
      }
      mx = 0;
      col = 0;
      last = a[i];
      idx++;
      continue;
    }
    mx = max(mx, a[i]);
    col++;
  }
  if (x < y * k) {
    if (col / k == 0) {
      if (mx > last) {
        cout << -1 << endl;
        return 0;
      }
      ans += (col * y);
    } else {
      long long pol = col / k, ost = (col % k);
      ans += (ost * y);
      ans += (pol * x);
    }
  }
  if (x == y * k) {
    if (col / k == 0) {
      if (mx > last) {
        cout << -1 << endl;
        return 0;
      }
      ans += (col * y);
    } else {
      long long pol = col / k, ost = (col % k);
      ans += (ost * y);
      ans += (pol * x);
    }
  }
  if (x > y * k) {
    if (col / k == 0) {
      if (mx > last) {
        cout << -1 << endl;
        return 0;
      }
      ans += (col * y);
    } else {
      if (mx > last) {
        long long ost = col - k;
        ans += x;
        ans += (ost * y);
      } else {
        ans += (col * y);
      }
    }
  }
  if (idx != m + 1) ans = -1;
  cout << ans << endl;
}

```