#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 5;
long long n, m, x, k, y, a[N], b[N];
long long destroy(long long l, long long r) {
  long long s = r - l, ma = *max_element(a + l, a + r);
  bool al = (l && a[l - 1] > ma) || (r < n && a[r] > ma);
  if (x > y * k) {
    if (al) return y * s;
    if (k > s) return -1;
    return x + (s - k) * y;
  }
  if (al || s >= k) return s / k * x + s % k * y;
  return -1;
}
long long mana() {
  long long j = 0, s = -1, ma = 0;
  for (long long i = 0; i < n; i++)
    if (j < m && a[i] == b[j]) {
      if (s != -1) {
        long long d = destroy(s, i);
        if (d == -1) return -1;
        ma += d;
        s = -1;
      }
      j++;
    } else if (s == -1)
      s = i;
  if (s != -1) {
    long long d = destroy(s, n);
    if (d == -1) return -1;
    ma += d;
  }
  return j == m ? ma : -1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> 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];
  cout << mana() << '\n';
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
const long long N = 2e5 + 5;
long long n, m, x, k, y, a[N], b[N];
long long destroy(long long l, long long r) {
  long long s = r - l, ma = *max_element(a + l, a + r);
  bool al = (l && a[l - 1] > ma) || (r < n && a[r] > ma);
  if (x > y * k) {
    if (al) return y * s;
    if (k > s) return -1;
    return x + (s - k) * y;
  }
  if (al || s >= k) return s / k * x + s % k * y;
  return -1;
}
long long mana() {
  long long j = 0, s = -1, ma = 0;
  for (long long i = 0; i < n; i++)
    if (j < m && a[i] == b[j]) {
      if (s != -1) {
        long long d = destroy(s, i);
        if (d == -1) return -1;
        ma += d;
        s = -1;
      }
      j++;
    } else if (s == -1)
      s = i;
  if (s != -1) {
    long long d = destroy(s, n);
    if (d == -1) return -1;
    ma += d;
  }
  return j == m ? ma : -1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> 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];
  cout << mana() << '\n';
}

```