#include <bits/stdc++.h>
using namespace std;
using ll = unsigned long long;
int N, M;
ll X, K, Y;
int main() {
  cin >> N >> M >> X >> K >> Y;
  vector<int> A(N);
  vector<int> B(M);
  map<int, int> A_ri;
  int max_idx_A = 0;
  for (int i = 0; i < N; i++) {
    cin >> A[i];
    A_ri[A[i]] = i;
    if (A[i] > A[max_idx_A]) {
      max_idx_A = i;
    }
  }
  for (int i = 0; i < M; i++) cin >> B[i];
  int last_idx = -1;
  set<int> A_idx_keep;
  vector<int> A_idx_keep_vec;
  for (int i = 0; i < M; i++) {
    if (A_ri.find(B[i]) == A_ri.end()) {
      cout << -1 << endl;
      return 0;
    } else {
      if (A_ri[B[i]] < last_idx) {
        cout << -1 << endl;
        return 0;
      } else {
        A_idx_keep.insert(A_ri[B[i]]);
        A_idx_keep_vec.push_back(A_ri[B[i]]);
        last_idx = A_ri[B[i]];
      }
    }
  }
  auto cost_to_destroy = [&](vector<int> segment, int left, int right) {
    if (segment.size() == 0) return (ll)0;
    ll seg_max = *max_element(segment.begin(), segment.end());
    if (segment.size() < K) {
      if (max(left, right) < seg_max) {
        cout << -1 << endl;
        exit(0);
      } else {
        return (ll)segment.size() * Y;
      }
    }
    if (Y * K < X) {
      if (max(left, right) > seg_max) {
        return (ll)segment.size() * Y;
      }
      return (ll)(segment.size() - K) * Y + X;
    }
    return (ll)(segment.size() % K) * Y + (segment.size() / K) * X;
  };
  ll ans = 0;
  A_idx_keep_vec.push_back(N);
  bool max_ok = false;
  for (int i = 0; i < A_idx_keep_vec.size(); i++) {
    int left = -1;
    int right = A_idx_keep_vec[i];
    vector<int> segment;
    if (i > 0) left = A_idx_keep_vec[i - 1];
    for (int j = left + 1; j < right; j++) {
      segment.emplace_back(A[j]);
    }
    int left_boundary = -1;
    if (left != -1) left_boundary = A[left];
    int right_boundary = -1;
    if (right != N) right_boundary = A[right];
    ans += cost_to_destroy(segment, left_boundary, right_boundary);
  }
  cout << ans << endl;
}

### 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;
using ll = unsigned long long;
int N, M;
ll X, K, Y;
int main() {
  cin >> N >> M >> X >> K >> Y;
  vector<int> A(N);
  vector<int> B(M);
  map<int, int> A_ri;
  int max_idx_A = 0;
  for (int i = 0; i < N; i++) {
    cin >> A[i];
    A_ri[A[i]] = i;
    if (A[i] > A[max_idx_A]) {
      max_idx_A = i;
    }
  }
  for (int i = 0; i < M; i++) cin >> B[i];
  int last_idx = -1;
  set<int> A_idx_keep;
  vector<int> A_idx_keep_vec;
  for (int i = 0; i < M; i++) {
    if (A_ri.find(B[i]) == A_ri.end()) {
      cout << -1 << endl;
      return 0;
    } else {
      if (A_ri[B[i]] < last_idx) {
        cout << -1 << endl;
        return 0;
      } else {
        A_idx_keep.insert(A_ri[B[i]]);
        A_idx_keep_vec.push_back(A_ri[B[i]]);
        last_idx = A_ri[B[i]];
      }
    }
  }
  auto cost_to_destroy = [&](vector<int> segment, int left, int right) {
    if (segment.size() == 0) return (ll)0;
    ll seg_max = *max_element(segment.begin(), segment.end());
    if (segment.size() < K) {
      if (max(left, right) < seg_max) {
        cout << -1 << endl;
        exit(0);
      } else {
        return (ll)segment.size() * Y;
      }
    }
    if (Y * K < X) {
      if (max(left, right) > seg_max) {
        return (ll)segment.size() * Y;
      }
      return (ll)(segment.size() - K) * Y + X;
    }
    return (ll)(segment.size() % K) * Y + (segment.size() / K) * X;
  };
  ll ans = 0;
  A_idx_keep_vec.push_back(N);
  bool max_ok = false;
  for (int i = 0; i < A_idx_keep_vec.size(); i++) {
    int left = -1;
    int right = A_idx_keep_vec[i];
    vector<int> segment;
    if (i > 0) left = A_idx_keep_vec[i - 1];
    for (int j = left + 1; j < right; j++) {
      segment.emplace_back(A[j]);
    }
    int left_boundary = -1;
    if (left != -1) left_boundary = A[left];
    int right_boundary = -1;
    if (right != N) right_boundary = A[right];
    ans += cost_to_destroy(segment, left_boundary, right_boundary);
  }
  cout << ans << endl;
}

```