#include <bits/stdc++.h>
using ll = long long;
using namespace std;
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(15);
  ll n, m, x, k, y;
  cin >> n >> m >> x >> k >> y;
  vector<ll> a(n + 1), b(m + 2);
  for (ll i = 0, iLen = (n); i < iLen; ++i) cin >> a[i];
  for (ll i = 1, iLen = (m + 1); i < iLen; ++i) cin >> b[i];
  a[n] = -1;
  b[m + 1] = -1;
  ll j = 1, left = 0, max_ai = 0;
  ll ans = 0;
  for (ll i = 0, iLen = (n + 1); i < iLen; ++i) {
    if (a[i] == b[j]) {
      if (left == 0) {
        max_ai = 0;
        left = 0;
        ++j;
        continue;
      }
      if (left > 0 && left < k && max_ai > max(b[j - 1], b[j])) {
        cout << -1 << endl;
        return 0;
      }
      if (max_ai > max(b[j - 1], b[j])) {
        left -= k;
        ans += x;
      }
      if (left % k) {
        ans += y * (left % k);
        left -= left % k;
      }
      ans += min(x * (left / k), y * left);
      left = 0;
      max_ai = 0;
      ++j;
    } else {
      max_ai = max(max_ai, a[i]);
      ++left;
    }
  }
  if (j != m + 2) {
    cout << -1 << endl;
  } else {
    cout << ans << endl;
  }
}

### Prompt

Please create a solution in cpp 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 ll = long long;
using namespace std;
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(15);
  ll n, m, x, k, y;
  cin >> n >> m >> x >> k >> y;
  vector<ll> a(n + 1), b(m + 2);
  for (ll i = 0, iLen = (n); i < iLen; ++i) cin >> a[i];
  for (ll i = 1, iLen = (m + 1); i < iLen; ++i) cin >> b[i];
  a[n] = -1;
  b[m + 1] = -1;
  ll j = 1, left = 0, max_ai = 0;
  ll ans = 0;
  for (ll i = 0, iLen = (n + 1); i < iLen; ++i) {
    if (a[i] == b[j]) {
      if (left == 0) {
        max_ai = 0;
        left = 0;
        ++j;
        continue;
      }
      if (left > 0 && left < k && max_ai > max(b[j - 1], b[j])) {
        cout << -1 << endl;
        return 0;
      }
      if (max_ai > max(b[j - 1], b[j])) {
        left -= k;
        ans += x;
      }
      if (left % k) {
        ans += y * (left % k);
        left -= left % k;
      }
      ans += min(x * (left / k), y * left);
      left = 0;
      max_ai = 0;
      ++j;
    } else {
      max_ai = max(max_ai, a[i]);
      ++left;
    }
  }
  if (j != m + 2) {
    cout << -1 << endl;
  } else {
    cout << ans << endl;
  }
}

```