#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7, N = 5e5 + 10;
const long long LINF = LLONG_MAX;
long long v[N];
long long b[N];
long long n, m, x, k, y;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m;
  cin >> x >> k >> y;
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
  }
  for (long long i = 0; i < m; i++) {
    cin >> b[i];
  }
  long long ans = 0;
  long long i = 0, j = 0, lst = -1;
  auto check = [&](long long l, long long r, long long &ans) -> bool {
    long long tmp = 0;
    if (l > r) {
      return true;
    }
    bool ch = false;
    long long mx = *max_element(v + l, v + r + 1);
    long long len = r - l + 1;
    if (l - 1 >= 0 and v[l - 1] > mx) {
      ch = true;
    }
    if (r + 1 < n and v[r + 1] > mx) {
      ch = true;
    }
    if (len < k and ch == false) {
      return false;
    }
    long long need = len % k;
    ans += need * y;
    len -= need;
    if (y * k >= x) {
      ans += (len / k) * x;
    } else if (ch) {
      ans += (len)*y;
    } else {
      ans += (len - k) * y + x;
    }
    return true;
  };
  while (j < m) {
    while (i < n and v[i] != b[j]) {
      i++;
    }
    if (i == n) {
      cout << -1 << "\n";
      return 0;
    }
    if (!check(lst + 1, i - 1, ans)) {
      cout << "-1\n";
      return 0;
    }
    lst = i;
    ++j;
  }
  if (!check(lst + 1, n - 1, ans)) {
    cout << "-1\n";
    return 0;
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to 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;
const long long MOD = 1e9 + 7, N = 5e5 + 10;
const long long LINF = LLONG_MAX;
long long v[N];
long long b[N];
long long n, m, x, k, y;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m;
  cin >> x >> k >> y;
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
  }
  for (long long i = 0; i < m; i++) {
    cin >> b[i];
  }
  long long ans = 0;
  long long i = 0, j = 0, lst = -1;
  auto check = [&](long long l, long long r, long long &ans) -> bool {
    long long tmp = 0;
    if (l > r) {
      return true;
    }
    bool ch = false;
    long long mx = *max_element(v + l, v + r + 1);
    long long len = r - l + 1;
    if (l - 1 >= 0 and v[l - 1] > mx) {
      ch = true;
    }
    if (r + 1 < n and v[r + 1] > mx) {
      ch = true;
    }
    if (len < k and ch == false) {
      return false;
    }
    long long need = len % k;
    ans += need * y;
    len -= need;
    if (y * k >= x) {
      ans += (len / k) * x;
    } else if (ch) {
      ans += (len)*y;
    } else {
      ans += (len - k) * y + x;
    }
    return true;
  };
  while (j < m) {
    while (i < n and v[i] != b[j]) {
      i++;
    }
    if (i == n) {
      cout << -1 << "\n";
      return 0;
    }
    if (!check(lst + 1, i - 1, ans)) {
      cout << "-1\n";
      return 0;
    }
    lst = i;
    ++j;
  }
  if (!check(lst + 1, n - 1, ans)) {
    cout << "-1\n";
    return 0;
  }
  cout << ans << "\n";
  return 0;
}

```