#include <bits/stdc++.h>
using namespace std;
long long getMaxNear(vector<long long>& b, long long index, long long m) {
  if (index == 0) return b[0];
  if (index == m) return b[m - 1];
  return max(b[index - 1], b[index]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  long long x, k, y;
  cin >> x >> k >> y;
  vector<long long> a(n), b(m);
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < m; ++i) cin >> b[i];
  vector<vector<long long>> vec(m + 1);
  int pos = 0;
  for (int i = 0; i < m; ++i) {
    long long maximum = -1, cnt = 0;
    while (pos < n && a[pos] != b[i]) {
      if (a[pos] > maximum) maximum = a[pos];
      cnt += 1;
      ++pos;
    }
    if (pos == n) {
      cout << -1 << "\n";
      return 0;
    }
    vec[i] = {cnt, maximum};
    ++pos;
  }
  long long maximum = -1, cnt = 0;
  while (pos < n) {
    if (a[pos] > maximum) maximum = a[pos];
    cnt += 1;
    ++pos;
  }
  vec[m] = {cnt, maximum};
  long long ans = 0, iter = 0;
  for (auto elem : vec) {
    if (x <= k * y) {
      if (getMaxNear(b, iter, m) < elem[1] && elem[0] < k) {
        cout << -1 << "\n";
        return 0;
      }
      ans += (elem[0] % k) * y;
      ans += (elem[0] / k) * x;
    } else {
      long long Maxnear = getMaxNear(b, iter, m);
      if (Maxnear > elem[1]) {
        ans += y * elem[0];
      } else {
        if (elem[0] < k) {
          cout << -1 << "\n";
          return 0;
        } else {
          ans += (x + (elem[0] - k) * y);
        }
      }
    }
    iter += 1;
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 getMaxNear(vector<long long>& b, long long index, long long m) {
  if (index == 0) return b[0];
  if (index == m) return b[m - 1];
  return max(b[index - 1], b[index]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  long long x, k, y;
  cin >> x >> k >> y;
  vector<long long> a(n), b(m);
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < m; ++i) cin >> b[i];
  vector<vector<long long>> vec(m + 1);
  int pos = 0;
  for (int i = 0; i < m; ++i) {
    long long maximum = -1, cnt = 0;
    while (pos < n && a[pos] != b[i]) {
      if (a[pos] > maximum) maximum = a[pos];
      cnt += 1;
      ++pos;
    }
    if (pos == n) {
      cout << -1 << "\n";
      return 0;
    }
    vec[i] = {cnt, maximum};
    ++pos;
  }
  long long maximum = -1, cnt = 0;
  while (pos < n) {
    if (a[pos] > maximum) maximum = a[pos];
    cnt += 1;
    ++pos;
  }
  vec[m] = {cnt, maximum};
  long long ans = 0, iter = 0;
  for (auto elem : vec) {
    if (x <= k * y) {
      if (getMaxNear(b, iter, m) < elem[1] && elem[0] < k) {
        cout << -1 << "\n";
        return 0;
      }
      ans += (elem[0] % k) * y;
      ans += (elem[0] / k) * x;
    } else {
      long long Maxnear = getMaxNear(b, iter, m);
      if (Maxnear > elem[1]) {
        ans += y * elem[0];
      } else {
        if (elem[0] < k) {
          cout << -1 << "\n";
          return 0;
        } else {
          ans += (x + (elem[0] - k) * y);
        }
      }
    }
    iter += 1;
  }
  cout << ans << "\n";
  return 0;
}

```