#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, m;
  cin >> n >> m;
  long long x, k, y;
  cin >> 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];
  vector<long long> t;
  long long answer = 0;
  long long left = 0, right = b[1], right_index = 1;
  for (long long i = 1; i <= n + 1; ++i) {
    if (a[i] == right) {
      long long length = t.size();
      if (length != 0) {
        long long berserks, fireballs;
        if (x <= k * y) {
          fireballs = length / k;
          berserks = length % k;
        } else {
          fireballs = 0;
          berserks = length;
        }
        long long maximal = *max_element(t.begin(), t.end());
        if (maximal > left && maximal > right) {
          if (length < k) {
            cout << "-1\n";
            return 0;
          }
          if (fireballs == 0) {
            fireballs += 1;
            berserks -= k;
          }
        }
        answer += fireballs * x + berserks * y;
      }
      t.clear();
      left = right;
      if (right_index != m + 1) right_index += 1;
      right = b[right_index];
    } else
      t.push_back(a[i]);
  }
  if (right != 0)
    cout << "-1\n";
  else
    cout << answer << "\n";
}

### Prompt

Your challenge is to write 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;
int main() {
  long long n, m;
  cin >> n >> m;
  long long x, k, y;
  cin >> 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];
  vector<long long> t;
  long long answer = 0;
  long long left = 0, right = b[1], right_index = 1;
  for (long long i = 1; i <= n + 1; ++i) {
    if (a[i] == right) {
      long long length = t.size();
      if (length != 0) {
        long long berserks, fireballs;
        if (x <= k * y) {
          fireballs = length / k;
          berserks = length % k;
        } else {
          fireballs = 0;
          berserks = length;
        }
        long long maximal = *max_element(t.begin(), t.end());
        if (maximal > left && maximal > right) {
          if (length < k) {
            cout << "-1\n";
            return 0;
          }
          if (fireballs == 0) {
            fireballs += 1;
            berserks -= k;
          }
        }
        answer += fireballs * x + berserks * y;
      }
      t.clear();
      left = right;
      if (right_index != m + 1) right_index += 1;
      right = b[right_index];
    } else
      t.push_back(a[i]);
  }
  if (right != 0)
    cout << "-1\n";
  else
    cout << answer << "\n";
}

```