#include <bits/stdc++.h>
using namespace std;
using ll = long long;
ll n, m, x, y, k, ans = 0;
ll solve(vector<int> c, int mx) {
  if (c.empty()) return 0;
  int m2 = *max_element(c.begin(), c.end());
  ll tans = 1ll << 60;
  for (int rem = 0; rem <= c.size(); rem++) {
    if (rem == c.size() && m2 > mx) continue;
    if ((c.size() - rem) % k) continue;
    ll cost = (c.size() - rem) / k;
    cost = cost * x + y * rem;
    tans = min(tans, cost);
  }
  if (tans == 1ll << 60) {
    cout << -1;
    exit(0);
  }
  return tans;
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin >> n >> m >> x >> k >> y;
  vector<int> a(n), b(m);
  for (auto &i : a) cin >> i;
  for (auto &i : b) cin >> i;
  a.push_back(-(1 << 30)), b.push_back(-(1 << 30)), n++, m++;
  int lst = -1;
  for (auto X : b) {
    int i = lst + 1;
    vector<int> c;
    while (i < n && a[i] != X) c.push_back(a[i++]);
    if (i == n) return cout << -1, 0;
    ans += solve(c, max(a[i], lst == -1 ? -1 : a[lst]));
    lst = i;
  }
  cout << ans << endl;
}

### Prompt

In Cpp, your task is to solve 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;
using ll = long long;
ll n, m, x, y, k, ans = 0;
ll solve(vector<int> c, int mx) {
  if (c.empty()) return 0;
  int m2 = *max_element(c.begin(), c.end());
  ll tans = 1ll << 60;
  for (int rem = 0; rem <= c.size(); rem++) {
    if (rem == c.size() && m2 > mx) continue;
    if ((c.size() - rem) % k) continue;
    ll cost = (c.size() - rem) / k;
    cost = cost * x + y * rem;
    tans = min(tans, cost);
  }
  if (tans == 1ll << 60) {
    cout << -1;
    exit(0);
  }
  return tans;
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin >> n >> m >> x >> k >> y;
  vector<int> a(n), b(m);
  for (auto &i : a) cin >> i;
  for (auto &i : b) cin >> i;
  a.push_back(-(1 << 30)), b.push_back(-(1 << 30)), n++, m++;
  int lst = -1;
  for (auto X : b) {
    int i = lst + 1;
    vector<int> c;
    while (i < n && a[i] != X) c.push_back(a[i++]);
    if (i == n) return cout << -1, 0;
    ans += solve(c, max(a[i], lst == -1 ? -1 : a[lst]));
    lst = i;
  }
  cout << ans << endl;
}

```