#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using VL = vector<ll>;
int n, m;
VL A, B;
int k;
ll x, y;
ll solu() {
  ll res = 0;
  int e = 0;
  for (int i = 0; i <= m; ++i) {
    int d = e;
    ll mx = 0;
    while (++d <= n + 1 and A[d] != B[i + 1]) mx = max(mx, A[d]);
    if (d > n + 1) return -1;
    ll mx2 = max(B[i], B[i + 1]);
    int s = d - e - 1;
    if (s) {
      ll r1, r2;
      if (mx2 > mx) {
        r1 = s * y;
        r2 = s / k * x + s % k * y;
      } else {
        if (s < k) return -1;
        r1 = (s - k) * y + x;
        r2 = s / k * x + s % k * y;
      }
      res += min(r1, r2);
    }
    e = d;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> x >> k >> y;
  A = VL(n + 2);
  B = VL(m + 2);
  for (int i = 1; i <= n; ++i) cin >> A[i];
  for (int i = 1; i <= m; ++i) cin >> B[i];
  cout << solu() << endl;
}

### Prompt

Develop a solution in cpp 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 namespace std;
using ll = long long;
using VL = vector<ll>;
int n, m;
VL A, B;
int k;
ll x, y;
ll solu() {
  ll res = 0;
  int e = 0;
  for (int i = 0; i <= m; ++i) {
    int d = e;
    ll mx = 0;
    while (++d <= n + 1 and A[d] != B[i + 1]) mx = max(mx, A[d]);
    if (d > n + 1) return -1;
    ll mx2 = max(B[i], B[i + 1]);
    int s = d - e - 1;
    if (s) {
      ll r1, r2;
      if (mx2 > mx) {
        r1 = s * y;
        r2 = s / k * x + s % k * y;
      } else {
        if (s < k) return -1;
        r1 = (s - k) * y + x;
        r2 = s / k * x + s % k * y;
      }
      res += min(r1, r2);
    }
    e = d;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> x >> k >> y;
  A = VL(n + 2);
  B = VL(m + 2);
  for (int i = 1; i <= n; ++i) cin >> A[i];
  for (int i = 1; i <= m; ++i) cin >> B[i];
  cout << solu() << endl;
}

```