#include <bits/stdc++.h>
using namespace std;
long long ans;
int n, m, a[200200], b[200200], id[200200], seg, k, one;
long long Solve(int L, int R) {
  if (L > R) return 0;
  int mx = *max_element(a + L, a + R + 1), Len = R - L + 1,
      len = (R - L + 1) % k;
  if (len == R - L + 1) {
    if (mx > a[L - 1] && mx > a[R + 1]) {
      cout << -1;
      exit(0);
    }
    return (len * (1LL) * one);
  }
  long long x, y, z = 0;
  Len -= len;
  z = len * 1LL * one;
  if (mx > a[L - 1] && mx > a[R + 1])
    x = (Len - k) * 1LL * one + seg;
  else
    x = Len * 1LL * one;
  y = (Len / k) * 1LL * seg;
  return min(x, y) + z;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> seg >> k >> one;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    id[a[i]] = i;
  }
  for (int i = 1; i <= m; i++) cin >> b[i];
  for (int i = 2; i <= m; i++)
    if (id[b[i]] < id[b[i - 1]]) return cout << -1, 0;
  ans = Solve(1, id[b[1]] - 1) + Solve(id[b[m]] + 1, n);
  for (int i = 1; i < m; i++) ans += Solve(id[b[i]] + 1, id[b[i + 1]] - 1);
  cout << ans;
  return 0;
}

### Prompt

Create a solution in cpp for 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;
long long ans;
int n, m, a[200200], b[200200], id[200200], seg, k, one;
long long Solve(int L, int R) {
  if (L > R) return 0;
  int mx = *max_element(a + L, a + R + 1), Len = R - L + 1,
      len = (R - L + 1) % k;
  if (len == R - L + 1) {
    if (mx > a[L - 1] && mx > a[R + 1]) {
      cout << -1;
      exit(0);
    }
    return (len * (1LL) * one);
  }
  long long x, y, z = 0;
  Len -= len;
  z = len * 1LL * one;
  if (mx > a[L - 1] && mx > a[R + 1])
    x = (Len - k) * 1LL * one + seg;
  else
    x = Len * 1LL * one;
  y = (Len / k) * 1LL * seg;
  return min(x, y) + z;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> seg >> k >> one;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    id[a[i]] = i;
  }
  for (int i = 1; i <= m; i++) cin >> b[i];
  for (int i = 2; i <= m; i++)
    if (id[b[i]] < id[b[i - 1]]) return cout << -1, 0;
  ans = Solve(1, id[b[1]] - 1) + Solve(id[b[m]] + 1, n);
  for (int i = 1; i < m; i++) ans += Solve(id[b[i]] + 1, id[b[i + 1]] - 1);
  cout << ans;
  return 0;
}

```