#include <bits/stdc++.h>
using namespace std;
long long first[200005];
long long second[200005];
int main(int argc, const char* argv[]) {
  long long n, m;
  cin >> n >> m;
  long long x, k, y;
  cin >> x >> k >> y;
  for (int i = 0; i < n; i++) {
    cin >> first[i];
  }
  for (int i = 0; i < m; i++) {
    cin >> second[i];
  }
  bool flag = false;
  if (n == m) {
    for (int i = 0; i < n; i++) {
      if (first[i] != second[i]) {
        flag = true;
        break;
      }
    }
    if (flag) {
      cout << -1 << endl;
      return 0;
    } else {
      cout << 0 << endl;
      return 0;
    }
  }
  flag = true;
  if (y * k > x) {
    flag = false;
  }
  long long ans = 0;
  if (flag) {
    long long head = 0;
    for (long long i = 0; i < m; i++) {
      long long tmax = -1;
      long long cnt = 0;
      for (long long j = head; j < n; j++) {
        if (first[j] == second[i]) {
          head = j + 1;
          break;
        } else {
          if (first[j] > tmax) {
            tmax = first[j];
          }
          cnt++;
        }
      }
      if (cnt) {
        if (tmax < second[i] || (i > 0 && tmax < second[i - 1])) {
          ans += cnt * y;
        } else {
          long long temp = cnt / k;
          if (temp == 0) {
            cout << -1 << endl;
            return 0;
          }
          ans += x;
          ans += y * (cnt - k);
        }
      }
    }
    if (head != n) {
      long long tmax = -1;
      long long cnt = 0;
      for (long long i = head; i < n; i++) {
        cnt++;
        if (first[i] > tmax) {
          tmax = first[i];
        }
      }
      if (tmax < second[m - 1]) {
        ans += cnt * y;
      } else {
        long long temp = cnt / k;
        if (temp == 0) {
          cout << -1 << endl;
          return 0;
        }
        ans += x;
        ans += y * (cnt - k);
      }
    }
  } else {
    long long head = 0;
    for (long long i = 0; i < m; i++) {
      long long tmax = -1;
      long long cnt = 0;
      for (long long j = head; j < n; j++) {
        if (first[j] == second[i]) {
          head = j + 1;
          break;
        } else {
          if (first[j] > tmax) {
            tmax = first[j];
          }
          cnt++;
        }
      }
      if (cnt) {
        long long temp = cnt / k;
        if (temp == 0) {
          if (tmax < second[i] || (i > 0 && tmax < second[i - 1])) {
            ans += cnt * y;
          } else {
            cout << -1 << endl;
            return 0;
          }
        } else {
          ans += temp * x;
          ans += y * (cnt % k);
        }
      }
    }
    if (head != n) {
      long long tmax = -1;
      long long cnt = 0;
      for (long long i = head; i < n; i++) {
        cnt++;
        if (first[i] > tmax) {
          tmax = first[i];
        }
      }
      long long temp = cnt / k;
      if (temp == 0) {
        if (tmax < second[m - 1]) {
          ans += cnt * y;
        } else {
          cout << -1 << endl;
          return 0;
        }
      } else {
        ans += temp * x;
        ans += y * (cnt % k);
      }
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Generate 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 first[200005];
long long second[200005];
int main(int argc, const char* argv[]) {
  long long n, m;
  cin >> n >> m;
  long long x, k, y;
  cin >> x >> k >> y;
  for (int i = 0; i < n; i++) {
    cin >> first[i];
  }
  for (int i = 0; i < m; i++) {
    cin >> second[i];
  }
  bool flag = false;
  if (n == m) {
    for (int i = 0; i < n; i++) {
      if (first[i] != second[i]) {
        flag = true;
        break;
      }
    }
    if (flag) {
      cout << -1 << endl;
      return 0;
    } else {
      cout << 0 << endl;
      return 0;
    }
  }
  flag = true;
  if (y * k > x) {
    flag = false;
  }
  long long ans = 0;
  if (flag) {
    long long head = 0;
    for (long long i = 0; i < m; i++) {
      long long tmax = -1;
      long long cnt = 0;
      for (long long j = head; j < n; j++) {
        if (first[j] == second[i]) {
          head = j + 1;
          break;
        } else {
          if (first[j] > tmax) {
            tmax = first[j];
          }
          cnt++;
        }
      }
      if (cnt) {
        if (tmax < second[i] || (i > 0 && tmax < second[i - 1])) {
          ans += cnt * y;
        } else {
          long long temp = cnt / k;
          if (temp == 0) {
            cout << -1 << endl;
            return 0;
          }
          ans += x;
          ans += y * (cnt - k);
        }
      }
    }
    if (head != n) {
      long long tmax = -1;
      long long cnt = 0;
      for (long long i = head; i < n; i++) {
        cnt++;
        if (first[i] > tmax) {
          tmax = first[i];
        }
      }
      if (tmax < second[m - 1]) {
        ans += cnt * y;
      } else {
        long long temp = cnt / k;
        if (temp == 0) {
          cout << -1 << endl;
          return 0;
        }
        ans += x;
        ans += y * (cnt - k);
      }
    }
  } else {
    long long head = 0;
    for (long long i = 0; i < m; i++) {
      long long tmax = -1;
      long long cnt = 0;
      for (long long j = head; j < n; j++) {
        if (first[j] == second[i]) {
          head = j + 1;
          break;
        } else {
          if (first[j] > tmax) {
            tmax = first[j];
          }
          cnt++;
        }
      }
      if (cnt) {
        long long temp = cnt / k;
        if (temp == 0) {
          if (tmax < second[i] || (i > 0 && tmax < second[i - 1])) {
            ans += cnt * y;
          } else {
            cout << -1 << endl;
            return 0;
          }
        } else {
          ans += temp * x;
          ans += y * (cnt % k);
        }
      }
    }
    if (head != n) {
      long long tmax = -1;
      long long cnt = 0;
      for (long long i = head; i < n; i++) {
        cnt++;
        if (first[i] > tmax) {
          tmax = first[i];
        }
      }
      long long temp = cnt / k;
      if (temp == 0) {
        if (tmax < second[m - 1]) {
          ans += cnt * y;
        } else {
          cout << -1 << endl;
          return 0;
        }
      } else {
        ans += temp * x;
        ans += y * (cnt % k);
      }
    }
  }
  cout << ans << endl;
  return 0;
}

```