#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b) {
  long long r = 1;
  while (b != 0) {
    if (b % 2 != 0) {
      r = r * a;
      r %= 998244353ll;
    }
    a = a * a;
    a = a % 998244353ll;
    b = b / 2;
  }
  return r;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    long long x, k, y, xp;
    cin >> x >> k >> y;
    bool sln = true;
    xp = min(x, y * k);
    long long a[n + 2];
    long long pos[n + 1];
    long long b[m + 1];
    for (long long i = (1); i <= (n); i++) {
      cin >> a[i];
      pos[a[i]] = i;
    }
    for (long long j = (1); j <= (m); j++) {
      cin >> b[j];
    }
    if (n < m) {
      cout << -1 << endl;
      return 0;
    }
    a[0] = 0;
    a[n + 1] = 0;
    long long pr = 0, ans = 0;
    for (long long i = (1); i <= (m + 1); i++) {
      long long req;
      if (i <= m) {
        req = pos[b[i]];
      } else {
        req = n + 1;
      }
      long long avl = req - pr - 1;
      if (avl < 0) {
        sln = false;
        break;
      }
      if (avl == 0) {
        pr = req;
        continue;
      }
      long long maxm = -1;
      long long psu = max(a[pr], a[req]);
      bool mxe = false;
      for (long long q = (pr + 1); q <= (req - 1); q++) {
        if (a[q] > psu) {
          mxe = true;
        }
      }
      if ((avl < k) && (mxe == true)) {
        sln = false;
        break;
      } else {
        ans += ((avl % k) * (y));
        avl = avl - (avl % k);
        if (mxe == true) {
          avl -= k;
          ans += x;
        }
        long long jsu = avl / k;
        ans += (jsu * xp);
      }
      pr = req;
    }
    if (sln) {
      cout << ans << endl;
    } else {
      cout << -1 << endl;
    }
  }
}

### Prompt

Create a solution in cpp for 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 power(long long a, long long b) {
  long long r = 1;
  while (b != 0) {
    if (b % 2 != 0) {
      r = r * a;
      r %= 998244353ll;
    }
    a = a * a;
    a = a % 998244353ll;
    b = b / 2;
  }
  return r;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    long long x, k, y, xp;
    cin >> x >> k >> y;
    bool sln = true;
    xp = min(x, y * k);
    long long a[n + 2];
    long long pos[n + 1];
    long long b[m + 1];
    for (long long i = (1); i <= (n); i++) {
      cin >> a[i];
      pos[a[i]] = i;
    }
    for (long long j = (1); j <= (m); j++) {
      cin >> b[j];
    }
    if (n < m) {
      cout << -1 << endl;
      return 0;
    }
    a[0] = 0;
    a[n + 1] = 0;
    long long pr = 0, ans = 0;
    for (long long i = (1); i <= (m + 1); i++) {
      long long req;
      if (i <= m) {
        req = pos[b[i]];
      } else {
        req = n + 1;
      }
      long long avl = req - pr - 1;
      if (avl < 0) {
        sln = false;
        break;
      }
      if (avl == 0) {
        pr = req;
        continue;
      }
      long long maxm = -1;
      long long psu = max(a[pr], a[req]);
      bool mxe = false;
      for (long long q = (pr + 1); q <= (req - 1); q++) {
        if (a[q] > psu) {
          mxe = true;
        }
      }
      if ((avl < k) && (mxe == true)) {
        sln = false;
        break;
      } else {
        ans += ((avl % k) * (y));
        avl = avl - (avl % k);
        if (mxe == true) {
          avl -= k;
          ans += x;
        }
        long long jsu = avl / k;
        ans += (jsu * xp);
      }
      pr = req;
    }
    if (sln) {
      cout << ans << endl;
    } else {
      cout << -1 << endl;
    }
  }
}

```