#include <bits/stdc++.h>
using namespace std;
const long long N = 5e6 + 5;
int main() {
  long long n, m;
  scanf("%lld", &n);
  ;
  scanf("%lld", &m);
  ;
  long long a, k, b;
  scanf("%lld", &a);
  ;
  scanf("%lld", &k);
  ;
  scanf("%lld", &b);
  ;
  long long arr[n], brr[m];
  map<long long, long long> mp;
  for (long long i = 0; i < n; i++) {
    scanf("%lld", &arr[i]);
    ;
    mp[arr[i]] = i;
  }
  long long cnt = -1;
  for (long long i = 0; i < m; i++) {
    scanf("%lld", &brr[i]);
    ;
    if (mp[brr[i]] > cnt)
      cnt = mp[brr[i]];
    else {
      cout << -1 << "\n";
      return 0;
    }
  }
  long long ans = 0;
  for (long long i = 0; i < m; i++) {
    if (i == 0) {
      if (mp[brr[i]] == 0)
        continue;
      else {
        long long l = mp[brr[i]];
        bool p = 0;
        for (long long j = 0; j < mp[brr[i]]; j++) {
          if (arr[j] > brr[i]) p = 1;
        }
        if (p == 1 && l < k) {
          cout << (-1) << "\n";
          return 0;
        } else if (p) {
          ans = ans + (long long)(l % k) * b;
          ans = ans + (long long)a;
          l = l - l % k;
          l = l - k;
          if (a <= k * b) {
            ans = ans + (l / k) * a;
          } else {
            ans = ans + l * b;
          }
        } else {
          ans = ans + (long long)(l % k) * b;
          l = l - l % k;
          if (a <= k * b) {
            ans = ans + (l / k) * a;
          } else {
            ans = ans + l * b;
          }
        }
      }
    } else {
      long long l = mp[brr[i]] - mp[brr[i - 1]] - 1;
      bool p = 0;
      for (long long j = mp[brr[i - 1]] + 1; j < mp[brr[i]]; j++) {
        if (arr[j] > brr[i] && arr[j] > brr[i - 1]) p = 1;
      }
      if (p == 1 && l < k) {
        cout << (-1) << "\n";
        return 0;
      } else if (p) {
        ans = ans + (long long)(l % k) * b;
        ans = ans + (long long)a;
        l = l - l % k;
        l = l - k;
        if (a <= k * b) {
          ans = ans + (l / k) * a;
        } else {
          ans = ans + l * b;
        }
      } else {
        ans = ans + (long long)(l % k) * b;
        l = l - l % k;
        if (a <= k * b) {
          ans = ans + (l / k) * a;
        } else {
          ans = ans + l * b;
        }
      }
    }
  }
  long long i = m - 1;
  if (mp[brr[i]] != n - 1) {
    long long l = n - mp[brr[i]] - 1;
    bool p = 0;
    for (long long j = mp[brr[i]]; j < n; j++) {
      if (arr[j] > brr[i]) p = 1;
    }
    if (p == 1 && l < k) {
      cout << (-1) << "\n";
      return 0;
    } else if (p) {
      ans = ans + (long long)(l % k) * b;
      ans = ans + (long long)a;
      l = l - l % k;
      l = l - k;
      if (a <= k * b) {
        ans = ans + (l / k) * a;
      } else {
        ans = ans + l * b;
      }
    } else {
      ans = ans + (long long)(l % k) * b;
      l = l - l % k;
      if (a <= k * b) {
        ans = ans + (l / k) * a;
      } else {
        ans = ans + l * b;
      }
    }
  }
  cout << ans << "\n";
}

### 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;
const long long N = 5e6 + 5;
int main() {
  long long n, m;
  scanf("%lld", &n);
  ;
  scanf("%lld", &m);
  ;
  long long a, k, b;
  scanf("%lld", &a);
  ;
  scanf("%lld", &k);
  ;
  scanf("%lld", &b);
  ;
  long long arr[n], brr[m];
  map<long long, long long> mp;
  for (long long i = 0; i < n; i++) {
    scanf("%lld", &arr[i]);
    ;
    mp[arr[i]] = i;
  }
  long long cnt = -1;
  for (long long i = 0; i < m; i++) {
    scanf("%lld", &brr[i]);
    ;
    if (mp[brr[i]] > cnt)
      cnt = mp[brr[i]];
    else {
      cout << -1 << "\n";
      return 0;
    }
  }
  long long ans = 0;
  for (long long i = 0; i < m; i++) {
    if (i == 0) {
      if (mp[brr[i]] == 0)
        continue;
      else {
        long long l = mp[brr[i]];
        bool p = 0;
        for (long long j = 0; j < mp[brr[i]]; j++) {
          if (arr[j] > brr[i]) p = 1;
        }
        if (p == 1 && l < k) {
          cout << (-1) << "\n";
          return 0;
        } else if (p) {
          ans = ans + (long long)(l % k) * b;
          ans = ans + (long long)a;
          l = l - l % k;
          l = l - k;
          if (a <= k * b) {
            ans = ans + (l / k) * a;
          } else {
            ans = ans + l * b;
          }
        } else {
          ans = ans + (long long)(l % k) * b;
          l = l - l % k;
          if (a <= k * b) {
            ans = ans + (l / k) * a;
          } else {
            ans = ans + l * b;
          }
        }
      }
    } else {
      long long l = mp[brr[i]] - mp[brr[i - 1]] - 1;
      bool p = 0;
      for (long long j = mp[brr[i - 1]] + 1; j < mp[brr[i]]; j++) {
        if (arr[j] > brr[i] && arr[j] > brr[i - 1]) p = 1;
      }
      if (p == 1 && l < k) {
        cout << (-1) << "\n";
        return 0;
      } else if (p) {
        ans = ans + (long long)(l % k) * b;
        ans = ans + (long long)a;
        l = l - l % k;
        l = l - k;
        if (a <= k * b) {
          ans = ans + (l / k) * a;
        } else {
          ans = ans + l * b;
        }
      } else {
        ans = ans + (long long)(l % k) * b;
        l = l - l % k;
        if (a <= k * b) {
          ans = ans + (l / k) * a;
        } else {
          ans = ans + l * b;
        }
      }
    }
  }
  long long i = m - 1;
  if (mp[brr[i]] != n - 1) {
    long long l = n - mp[brr[i]] - 1;
    bool p = 0;
    for (long long j = mp[brr[i]]; j < n; j++) {
      if (arr[j] > brr[i]) p = 1;
    }
    if (p == 1 && l < k) {
      cout << (-1) << "\n";
      return 0;
    } else if (p) {
      ans = ans + (long long)(l % k) * b;
      ans = ans + (long long)a;
      l = l - l % k;
      l = l - k;
      if (a <= k * b) {
        ans = ans + (l / k) * a;
      } else {
        ans = ans + l * b;
      }
    } else {
      ans = ans + (long long)(l % k) * b;
      l = l - l % k;
      if (a <= k * b) {
        ans = ans + (l / k) * a;
      } else {
        ans = ans + l * b;
      }
    }
  }
  cout << ans << "\n";
}

```