#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, m, x, y, k;
  cin >> n >> m;
  long long a[n], b[m];
  cin >> x >> k >> y;
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < m; i++) cin >> b[i];
  vector<long long> adj[n];
  long long z[n];
  long long ind = 0;
  for (long long i = 0; i < n; i++) {
    if (ind >= m) {
      adj[ind].push_back(a[i]);
      continue;
    }
    if (b[ind] == a[i]) {
      if (ind - 1 >= 0)
        z[ind] = max(b[ind], b[ind - 1]);
      else
        z[ind] = b[ind];
      ind++;
    } else
      adj[ind].push_back(a[i]);
  }
  if (ind < m - 1) {
    cout << "-1\n";
    return 0;
  }
  z[ind] = b[m - 1];
  ind = 0;
  long long ans = 0;
  int f = 0;
  for (auto &it : adj) {
    long long itl = it.size();
    if (itl == 0) {
      ind++;
      continue;
    }
    long long mxv = -1;
    for (auto &it2 : it) mxv = max(mxv, it2);
    if (itl < k) {
      if (mxv < z[ind])
        ans += itl * y;
      else {
        f = 1;
        break;
      }
    } else {
      if (x >= y * k) {
        if (mxv < z[ind])
          ans += itl * y;
        else {
          ans = ans + (itl - k) * y + x;
        }
      } else {
        ans += (itl / k) * x + (itl % k) * y;
      }
    }
    ind++;
  }
  if (f == 1)
    cout << "-1\n";
  else
    cout << ans << "\n";
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
int main() {
  long long n, m, x, y, k;
  cin >> n >> m;
  long long a[n], b[m];
  cin >> x >> k >> y;
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < m; i++) cin >> b[i];
  vector<long long> adj[n];
  long long z[n];
  long long ind = 0;
  for (long long i = 0; i < n; i++) {
    if (ind >= m) {
      adj[ind].push_back(a[i]);
      continue;
    }
    if (b[ind] == a[i]) {
      if (ind - 1 >= 0)
        z[ind] = max(b[ind], b[ind - 1]);
      else
        z[ind] = b[ind];
      ind++;
    } else
      adj[ind].push_back(a[i]);
  }
  if (ind < m - 1) {
    cout << "-1\n";
    return 0;
  }
  z[ind] = b[m - 1];
  ind = 0;
  long long ans = 0;
  int f = 0;
  for (auto &it : adj) {
    long long itl = it.size();
    if (itl == 0) {
      ind++;
      continue;
    }
    long long mxv = -1;
    for (auto &it2 : it) mxv = max(mxv, it2);
    if (itl < k) {
      if (mxv < z[ind])
        ans += itl * y;
      else {
        f = 1;
        break;
      }
    } else {
      if (x >= y * k) {
        if (mxv < z[ind])
          ans += itl * y;
        else {
          ans = ans + (itl - k) * y + x;
        }
      } else {
        ans += (itl / k) * x + (itl % k) * y;
      }
    }
    ind++;
  }
  if (f == 1)
    cout << "-1\n";
  else
    cout << ans << "\n";
  return 0;
}

```