#include <bits/stdc++.h>
using namespace std;
template <typename T>
void input(vector<T> &arr, long long int n) {
  T temp;
  for (long long int i = 0; i < n; i++) cin >> temp, arr.push_back(temp);
}
template <typename T>
void output(vector<T> arr) {
  for (auto x : arr) cout << x << " ";
  cout << endl;
}
template <typename T>
void input_set(set<T> &arr, long long int n) {
  T temp;
  for (long long int i = 0; i < n; i++) cin >> temp, arr.insert(temp);
}
long long int mul(long long int a, long long int b) {
  return (a % 1000000007 * b % 1000000007) % 1000000007;
}
long long int power(long long int a, long long int b) {
  long long int ans = 1;
  while (b > 0) {
    if (b & 1) ans = mul(ans, a);
    a = mul(a, a);
    ;
    b >>= 1;
  }
  return ans;
}
vector<long long int> a, b;
bool check() {
  map<long long int, long long int> make_pair;
  for (int i = 0; i < a.size(); i++) make_pair[a[i]] = i;
  int ma = -1;
  for (int i = 0; i < b.size(); i++) {
    if (make_pair[b[i]] < ma) return false;
    ma = make_pair[b[i]];
  }
  return true;
}
void solve() {
  int n, m;
  cin >> n >> m;
  long long int x, k, y;
  cin >> x >> k >> y;
  a.clear(), b.clear();
  input(a, n);
  input(b, m);
  bool possible = check();
  if (!possible) {
    cout << -1 << endl;
    return;
  }
  map<long long int, long long int> make_pair;
  for (int i = 0; i < a.size(); i++) make_pair[a[i]] = i;
  vector<bool> marked(n, true);
  for (auto x : b) marked[make_pair[x]] = false;
  long long int ans = 0;
  int i = 0;
  while (i < n) {
    int j = i;
    long long int ma = -1;
    if (!marked[i]) {
      ++i;
      continue;
    }
    while (j < n and marked[j]) {
      ma = max(ma, a[j]);
      ++j;
    }
    int cnt = j - i;
    long long int left = (i == 0) ? -1 : a[i - 1];
    long long int right = (j == n) ? -1 : a[j];
    long long int temp = 0;
    if (cnt >= k) {
      temp = (cnt / k) * x + (cnt % k) * y;
      if (ma < left or ma < right) {
        temp = min(temp, cnt * y);
      }
      temp = min(temp, x + (cnt - k) * y);
      ans += temp;
    } else {
      if (ma > left and ma > right) {
        possible = false;
        break;
      }
      ans += cnt * y;
    }
    i = j;
  }
  if (possible)
    cout << ans << endl;
  else
    cout << -1 << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int testcases = 1;
  while (testcases--) {
    solve();
  }
}

### 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;
template <typename T>
void input(vector<T> &arr, long long int n) {
  T temp;
  for (long long int i = 0; i < n; i++) cin >> temp, arr.push_back(temp);
}
template <typename T>
void output(vector<T> arr) {
  for (auto x : arr) cout << x << " ";
  cout << endl;
}
template <typename T>
void input_set(set<T> &arr, long long int n) {
  T temp;
  for (long long int i = 0; i < n; i++) cin >> temp, arr.insert(temp);
}
long long int mul(long long int a, long long int b) {
  return (a % 1000000007 * b % 1000000007) % 1000000007;
}
long long int power(long long int a, long long int b) {
  long long int ans = 1;
  while (b > 0) {
    if (b & 1) ans = mul(ans, a);
    a = mul(a, a);
    ;
    b >>= 1;
  }
  return ans;
}
vector<long long int> a, b;
bool check() {
  map<long long int, long long int> make_pair;
  for (int i = 0; i < a.size(); i++) make_pair[a[i]] = i;
  int ma = -1;
  for (int i = 0; i < b.size(); i++) {
    if (make_pair[b[i]] < ma) return false;
    ma = make_pair[b[i]];
  }
  return true;
}
void solve() {
  int n, m;
  cin >> n >> m;
  long long int x, k, y;
  cin >> x >> k >> y;
  a.clear(), b.clear();
  input(a, n);
  input(b, m);
  bool possible = check();
  if (!possible) {
    cout << -1 << endl;
    return;
  }
  map<long long int, long long int> make_pair;
  for (int i = 0; i < a.size(); i++) make_pair[a[i]] = i;
  vector<bool> marked(n, true);
  for (auto x : b) marked[make_pair[x]] = false;
  long long int ans = 0;
  int i = 0;
  while (i < n) {
    int j = i;
    long long int ma = -1;
    if (!marked[i]) {
      ++i;
      continue;
    }
    while (j < n and marked[j]) {
      ma = max(ma, a[j]);
      ++j;
    }
    int cnt = j - i;
    long long int left = (i == 0) ? -1 : a[i - 1];
    long long int right = (j == n) ? -1 : a[j];
    long long int temp = 0;
    if (cnt >= k) {
      temp = (cnt / k) * x + (cnt % k) * y;
      if (ma < left or ma < right) {
        temp = min(temp, cnt * y);
      }
      temp = min(temp, x + (cnt - k) * y);
      ans += temp;
    } else {
      if (ma > left and ma > right) {
        possible = false;
        break;
      }
      ans += cnt * y;
    }
    i = j;
  }
  if (possible)
    cout << ans << endl;
  else
    cout << -1 << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int testcases = 1;
  while (testcases--) {
    solve();
  }
}

```