#include <bits/stdc++.h>
using ll = long long;
using namespace std;
int a[200010], b[200010];
bool ex[200010];
int n, m;
int x, k, y;
ll calc(int l, int r) {
  if (l + 1 == r) return 0;
  int rng = r - l + 1 - 2;
  int mx = *max_element(a + l, a + r + 1);
  if (mx > a[l] && mx > a[r] && rng < k) {
    cout << -1;
    exit(0);
  }
  if (1LL * y * k < x) {
    if (mx > a[l] && mx > a[r]) {
      return 1LL * y * (rng - k) + x;
    } else {
      return 1LL * y * rng;
    }
  } else {
    return 1LL * rng / k * x + 1LL * y * (rng % k);
  }
}
int main() {
  cin >> n >> m;
  cin >> x >> k >> y;
  for (int i = 0; i < n; i++) cin >> a[i + 1];
  for (int i = 0; i < m; i++) {
    cin >> b[i + 1];
  }
  int pos = 1;
  for (int i = 1; i <= n; i++) {
    if (a[i] == b[pos]) ex[i] = 1, pos++;
  }
  if (pos != m + 1) return 0 * puts("-1");
  ll ans = 0;
  int l = 0;
  ex[n + 1] = 1;
  for (int i = 1; i <= n + 1; i++) {
    if (ex[i]) {
      ans += calc(l, i);
      l = i;
    }
  }
  cout << ans;
  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 ll = long long;
using namespace std;
int a[200010], b[200010];
bool ex[200010];
int n, m;
int x, k, y;
ll calc(int l, int r) {
  if (l + 1 == r) return 0;
  int rng = r - l + 1 - 2;
  int mx = *max_element(a + l, a + r + 1);
  if (mx > a[l] && mx > a[r] && rng < k) {
    cout << -1;
    exit(0);
  }
  if (1LL * y * k < x) {
    if (mx > a[l] && mx > a[r]) {
      return 1LL * y * (rng - k) + x;
    } else {
      return 1LL * y * rng;
    }
  } else {
    return 1LL * rng / k * x + 1LL * y * (rng % k);
  }
}
int main() {
  cin >> n >> m;
  cin >> x >> k >> y;
  for (int i = 0; i < n; i++) cin >> a[i + 1];
  for (int i = 0; i < m; i++) {
    cin >> b[i + 1];
  }
  int pos = 1;
  for (int i = 1; i <= n; i++) {
    if (a[i] == b[pos]) ex[i] = 1, pos++;
  }
  if (pos != m + 1) return 0 * puts("-1");
  ll ans = 0;
  int l = 0;
  ex[n + 1] = 1;
  for (int i = 1; i <= n + 1; i++) {
    if (ex[i]) {
      ans += calc(l, i);
      l = i;
    }
  }
  cout << ans;
  return 0;
}

```