#include <bits/stdc++.h>
using namespace std;
bool isFile = 0;
const long long INF = 1e15 + 1234;
const int N = 2e5 + 1234;
const long long MOD = 998244353;
const long long mod = 1e9 + 7;
double EPS = 3e-18;
const double PI = acos(-1);
void faster() {
  if (isFile) {
    freopen(
        "sort"
        ".in",
        "r", stdin);
    freopen(
        "sort"
        ".out",
        "w", stdout);
  }
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
void solve() {
  int n, m;
  long long x, k, y;
  std::cin >> n >> m >> x >> k >> y;
  std::vector<int> a(n), b(m);
  for (int& i : a) std::cin >> i;
  for (int& i : b) std::cin >> i;
  int cnt = 0, mx = 0, prv = -1, us = 0;
  long long ans = 0;
  a.push_back(0);
  b.push_back(0);
  for (int i = 0; i <= n; i++) {
    if (us <= m && a[i] == b[us]) {
      long long now = 1e15;
      if (mx < prv || mx < a[i]) now = y * cnt;
      for (int j = k; j <= cnt; j += k)
        now = std::min(now, x * (j / k) + y * (cnt - j));
      if (now >= 1e15) {
        std::cout << "-1\n";
        return;
      }
      ans += now;
      us++;
      prv = a[i];
      mx = 0;
      cnt = 0;
    } else
      mx = std::max(mx, a[i]), cnt++;
  }
  std::cout << (us == m + 1 ? ans : -1) << "\n";
}
int main() {
  faster();
  int t = 1;
  while (t--) {
    solve();
    cout << "\n";
  }
  return 0;
}

### 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;
bool isFile = 0;
const long long INF = 1e15 + 1234;
const int N = 2e5 + 1234;
const long long MOD = 998244353;
const long long mod = 1e9 + 7;
double EPS = 3e-18;
const double PI = acos(-1);
void faster() {
  if (isFile) {
    freopen(
        "sort"
        ".in",
        "r", stdin);
    freopen(
        "sort"
        ".out",
        "w", stdout);
  }
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
void solve() {
  int n, m;
  long long x, k, y;
  std::cin >> n >> m >> x >> k >> y;
  std::vector<int> a(n), b(m);
  for (int& i : a) std::cin >> i;
  for (int& i : b) std::cin >> i;
  int cnt = 0, mx = 0, prv = -1, us = 0;
  long long ans = 0;
  a.push_back(0);
  b.push_back(0);
  for (int i = 0; i <= n; i++) {
    if (us <= m && a[i] == b[us]) {
      long long now = 1e15;
      if (mx < prv || mx < a[i]) now = y * cnt;
      for (int j = k; j <= cnt; j += k)
        now = std::min(now, x * (j / k) + y * (cnt - j));
      if (now >= 1e15) {
        std::cout << "-1\n";
        return;
      }
      ans += now;
      us++;
      prv = a[i];
      mx = 0;
      cnt = 0;
    } else
      mx = std::max(mx, a[i]), cnt++;
  }
  std::cout << (us == m + 1 ? ans : -1) << "\n";
}
int main() {
  faster();
  int t = 1;
  while (t--) {
    solve();
    cout << "\n";
  }
  return 0;
}

```