#include <bits/stdc++.h>
using namespace std;
const long long int N = 200005;
long long int a[N], b[N];
long long int x, k, y, ans = 0;
void go(long long int l, long long int r) {
  long long int val = *max_element(a + l, a + r + 1);
  if (r - l + 1 < k) {
    if (val > max(a[l - 1], a[r + 1])) {
      cout << "-1\n", exit(0);
    } else {
      ans += (r - l + 1) * y;
    }
  } else if (k * y < x) {
    if (val > max(a[l - 1], a[r + 1])) {
      ans += (r - l + 1 - k) * y + x;
    } else {
      ans += (r - l + 1) * y;
    }
  } else {
    ans += (r - l + 1) % k * y + (r - l + 1) / k * x;
  }
}
void solve() {
  long long int i, j, n, m, cnt = 0, sum = 0;
  cin >> n >> m;
  cin >> x >> k >> y;
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (i = 0; i < m; i++) {
    cin >> b[i];
  }
  j = 0;
  for (i = 0; i < n; i++) {
    if (a[i] == b[j]) {
      j++;
    }
  }
  if (j != m) {
    cout << -1 << '\n';
    return;
  }
  j = 0;
  cnt = 0, ans = 0;
  for (i = 0; i < m; ++i) {
    while (j < n && a[j] != b[i]) {
      ++j;
    }
    if (j >= n) {
      cout << -1 << '\n';
      return;
    }
    go(cnt, j - 1);
    cnt = j + 1;
  }
  if (cnt < n) {
    go(cnt, n - 1);
  }
  cout << ans << '\n';
  return;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
  return 0;
}

### Prompt

Generate 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 int N = 200005;
long long int a[N], b[N];
long long int x, k, y, ans = 0;
void go(long long int l, long long int r) {
  long long int val = *max_element(a + l, a + r + 1);
  if (r - l + 1 < k) {
    if (val > max(a[l - 1], a[r + 1])) {
      cout << "-1\n", exit(0);
    } else {
      ans += (r - l + 1) * y;
    }
  } else if (k * y < x) {
    if (val > max(a[l - 1], a[r + 1])) {
      ans += (r - l + 1 - k) * y + x;
    } else {
      ans += (r - l + 1) * y;
    }
  } else {
    ans += (r - l + 1) % k * y + (r - l + 1) / k * x;
  }
}
void solve() {
  long long int i, j, n, m, cnt = 0, sum = 0;
  cin >> n >> m;
  cin >> x >> k >> y;
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (i = 0; i < m; i++) {
    cin >> b[i];
  }
  j = 0;
  for (i = 0; i < n; i++) {
    if (a[i] == b[j]) {
      j++;
    }
  }
  if (j != m) {
    cout << -1 << '\n';
    return;
  }
  j = 0;
  cnt = 0, ans = 0;
  for (i = 0; i < m; ++i) {
    while (j < n && a[j] != b[i]) {
      ++j;
    }
    if (j >= n) {
      cout << -1 << '\n';
      return;
    }
    go(cnt, j - 1);
    cnt = j + 1;
  }
  if (cnt < n) {
    go(cnt, n - 1);
  }
  cout << ans << '\n';
  return;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
  return 0;
}

```