#include <bits/stdc++.h>
using namespace std;
using namespace std::chrono;
long long int powD(long long int x, long long int y) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % (long long int)(1e9 + 7);
    x = (x * x) % (long long int)(1e9 + 7);
    y = y >> 1;
  }
  return res;
}
int MAX = 2e5 + 1;
long long int x, k, y;
long long int n, m;
vector<long long int> a(MAX), b(MAX);
long long int calcMinm(int l, int r) {
  long long int maxm = -1;
  if (l != -1) maxm = max(maxm, a[l]);
  if (r != n) maxm = max(maxm, a[r]);
  long long int sz = r - l - 1;
  long long int cnt1 = 0;
  for (int i = l + 1; i < r; ++i) {
    if (a[i] > maxm) cnt1++;
  }
  long long int sum = 0;
  if (sz < k && cnt1) return -1;
  if (x < y * k) {
    return (sz % k) * y + sz / k * x;
  } else {
    if (cnt1) {
      return x + (sz - k) * y;
    } else {
      return sz * y;
    }
  }
}
int main() {
  cin >> n >> m;
  cin >> x >> k >> y;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < m; ++i) {
    cin >> b[i];
  }
  int j = 0;
  vector<int> idx;
  idx.push_back(-1);
  for (int i = 0; i < n && j < m; i++) {
    if (a[i] == b[j]) {
      idx.push_back(i);
      j++;
    }
  }
  idx.push_back(n);
  if (idx.size() != m + 2) {
    cout << -1 << "\n";
    return 0;
  }
  long long int ans = 0;
  bool flag = 1;
  for (int i = 0; i < m + 1; ++i) {
    long long int qty = calcMinm(idx[i], idx[i + 1]);
    if (qty == -1) {
      flag = 0;
      break;
    }
    ans += calcMinm(idx[i], idx[i + 1]);
  }
  if (flag)
    cout << ans << "\n";
  else
    cout << -1 << "\n";
}

### 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;
using namespace std::chrono;
long long int powD(long long int x, long long int y) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % (long long int)(1e9 + 7);
    x = (x * x) % (long long int)(1e9 + 7);
    y = y >> 1;
  }
  return res;
}
int MAX = 2e5 + 1;
long long int x, k, y;
long long int n, m;
vector<long long int> a(MAX), b(MAX);
long long int calcMinm(int l, int r) {
  long long int maxm = -1;
  if (l != -1) maxm = max(maxm, a[l]);
  if (r != n) maxm = max(maxm, a[r]);
  long long int sz = r - l - 1;
  long long int cnt1 = 0;
  for (int i = l + 1; i < r; ++i) {
    if (a[i] > maxm) cnt1++;
  }
  long long int sum = 0;
  if (sz < k && cnt1) return -1;
  if (x < y * k) {
    return (sz % k) * y + sz / k * x;
  } else {
    if (cnt1) {
      return x + (sz - k) * y;
    } else {
      return sz * y;
    }
  }
}
int main() {
  cin >> n >> m;
  cin >> x >> k >> y;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < m; ++i) {
    cin >> b[i];
  }
  int j = 0;
  vector<int> idx;
  idx.push_back(-1);
  for (int i = 0; i < n && j < m; i++) {
    if (a[i] == b[j]) {
      idx.push_back(i);
      j++;
    }
  }
  idx.push_back(n);
  if (idx.size() != m + 2) {
    cout << -1 << "\n";
    return 0;
  }
  long long int ans = 0;
  bool flag = 1;
  for (int i = 0; i < m + 1; ++i) {
    long long int qty = calcMinm(idx[i], idx[i + 1]);
    if (qty == -1) {
      flag = 0;
      break;
    }
    ans += calcMinm(idx[i], idx[i + 1]);
  }
  if (flag)
    cout << ans << "\n";
  else
    cout << -1 << "\n";
}

```