#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long T = 1;
  while (T--) {
    long long n, m, x, k, y;
    cin >> n >> m >> x >> k >> y;
    vector<long long> a(n + 2, 0), b(m + 2, 0);
    for (long long i = 1; i <= n; i++) cin >> a[i];
    for (long long i = 1; i <= m; i++) cin >> b[i];
    long long j = 1;
    vector<long long> v;
    v.push_back(0);
    for (long long i = 1; i <= n; i++) {
      if (a[i] == b[j]) {
        v.push_back(i);
        j++;
      }
    }
    v.push_back(n + 1);
    if (j != m + 1) {
      cout << "-1\n";
      return 0;
    }
    long long ans = 0;
    for (long long i = 1; i < (long long)v.size(); i++) {
      long long l = v[i - 1], r = v[i];
      long long mx = max(a[l], a[r]);
      vector<long long> temp;
      for (long long j = l + 1; j < r; j++) temp.push_back(a[j]);
      sort(temp.begin(), temp.end());
      long long sz = temp.size();
      if (sz > 0 && sz < k && temp[sz - 1] > mx) {
        cout << "-1\n";
        return 0;
      }
      if (sz > 0 && temp[sz - 1] > mx) {
        ans += x;
        sz -= k;
        for (long long i = 0; i < k; i++) temp.pop_back();
      }
      if (sz > 0) {
        long long ux = x * (sz / k) + y * (sz % k);
        long long uy = y * sz;
        ans += min(uy, ux);
      }
    }
    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;
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long T = 1;
  while (T--) {
    long long n, m, x, k, y;
    cin >> n >> m >> x >> k >> y;
    vector<long long> a(n + 2, 0), b(m + 2, 0);
    for (long long i = 1; i <= n; i++) cin >> a[i];
    for (long long i = 1; i <= m; i++) cin >> b[i];
    long long j = 1;
    vector<long long> v;
    v.push_back(0);
    for (long long i = 1; i <= n; i++) {
      if (a[i] == b[j]) {
        v.push_back(i);
        j++;
      }
    }
    v.push_back(n + 1);
    if (j != m + 1) {
      cout << "-1\n";
      return 0;
    }
    long long ans = 0;
    for (long long i = 1; i < (long long)v.size(); i++) {
      long long l = v[i - 1], r = v[i];
      long long mx = max(a[l], a[r]);
      vector<long long> temp;
      for (long long j = l + 1; j < r; j++) temp.push_back(a[j]);
      sort(temp.begin(), temp.end());
      long long sz = temp.size();
      if (sz > 0 && sz < k && temp[sz - 1] > mx) {
        cout << "-1\n";
        return 0;
      }
      if (sz > 0 && temp[sz - 1] > mx) {
        ans += x;
        sz -= k;
        for (long long i = 0; i < k; i++) temp.pop_back();
      }
      if (sz > 0) {
        long long ux = x * (sz / k) + y * (sz % k);
        long long uy = y * sz;
        ans += min(uy, ux);
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

```