#include <bits/stdc++.h>
using namespace std;
int cmp_double(double a, double b, double eps = 1e-9) {
  return a + eps > b ? b + eps > a ? 0 : 1 : -1;
}
bool canDo(const vector<int>& orig, const vector<int>& result) {
  int j = 0;
  for (int i = 0; i < result.size(); i++) {
    bool found = false;
    for (; j < orig.size(); j++) {
      if (orig[j] == result[i]) {
        j++;
        found = true;
        break;
      }
    }
    if (!found) return false;
  }
  return true;
}
int bigger_in_range(int i, int j, const vector<int>& orig, int guy) {
  int ans = 0;
  for (int k = i; k <= j; k++) {
    if (orig[k] > guy) ans++;
  }
  return ans;
}
long long computeAns(int kill_left, int kill_right, const vector<int>& orig,
                     int left_guy, int right_guy, int X, int K, int Y) {
  if (kill_left > kill_right) return 0;
  float mana_range = (float)X / (float)K;
  float single_kill = (float)Y;
  int range = (kill_right - kill_left) + 1;
  bool range_is_better = cmp_double(mana_range, single_kill) == -1;
  bool can_use_range = range >= K;
  int big_guy = max(left_guy, right_guy);
  int bigger_count = bigger_in_range(kill_left, kill_right, orig, big_guy);
  if (!can_use_range) {
    if (bigger_count > 0) {
      return -1;
    }
    return (long long)range * (long long)Y;
  }
  if (range_is_better) {
    int range_times = range / K;
    long long single_times =
        (long long)range - ((long long)range_times * (long long)K);
    return (long long)range_times * (long long)X +
           (long long)single_times * (long long)Y;
  }
  if (bigger_count > 0) {
    int range_times = 1;
    long long single_times =
        (long long)range - ((long long)range_times * (long long)K);
    return (long long)range_times * (long long)X +
           (long long)single_times * (long long)Y;
  }
  return (long long)range * (long long)Y;
}
int main() {
  int N, M;
  cin >> N >> M;
  int X, K, Y;
  cin >> X >> K >> Y;
  map<int, int> elem;
  vector<int> orig(N);
  for (int i = 0; i < N; i++) {
    cin >> orig[i];
    elem[orig[i]] = i;
  }
  vector<int> result(M);
  for (int i = 0; i < M; i++) {
    cin >> result[i];
  }
  if (!canDo(orig, result)) {
    cout << "-1" << endl;
    return 0;
  }
  vector<int> kept_pos;
  for (int i = 0; i < M; i++) {
    kept_pos.push_back(elem[result[i]]);
  }
  long long ans = 0;
  if (kept_pos[0] != 0) {
    long long aux =
        computeAns(0, kept_pos[0] - 1, orig, -1, result[0], X, K, Y);
    if (aux == -1) {
      cout << "-1" << endl;
      return 0;
    }
    ans += aux;
  }
  if (kept_pos[M - 1] != N - 1) {
    long long aux = computeAns(kept_pos[M - 1] + 1, N - 1, orig, result[M - 1],
                               -1, X, K, Y);
    if (aux == -1) {
      cout << "-1" << endl;
      return 0;
    }
    ans += aux;
  }
  if (M > 1) {
    for (int i = 1; i < M; i++) {
      long long aux = computeAns(kept_pos[i - 1] + 1, kept_pos[i] - 1, orig,
                                 result[i - 1], result[i], X, K, Y);
      if (aux == -1) {
        cout << "-1" << endl;
        return 0;
      }
      ans += aux;
    }
  }
  cout << ans << endl;
  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;
int cmp_double(double a, double b, double eps = 1e-9) {
  return a + eps > b ? b + eps > a ? 0 : 1 : -1;
}
bool canDo(const vector<int>& orig, const vector<int>& result) {
  int j = 0;
  for (int i = 0; i < result.size(); i++) {
    bool found = false;
    for (; j < orig.size(); j++) {
      if (orig[j] == result[i]) {
        j++;
        found = true;
        break;
      }
    }
    if (!found) return false;
  }
  return true;
}
int bigger_in_range(int i, int j, const vector<int>& orig, int guy) {
  int ans = 0;
  for (int k = i; k <= j; k++) {
    if (orig[k] > guy) ans++;
  }
  return ans;
}
long long computeAns(int kill_left, int kill_right, const vector<int>& orig,
                     int left_guy, int right_guy, int X, int K, int Y) {
  if (kill_left > kill_right) return 0;
  float mana_range = (float)X / (float)K;
  float single_kill = (float)Y;
  int range = (kill_right - kill_left) + 1;
  bool range_is_better = cmp_double(mana_range, single_kill) == -1;
  bool can_use_range = range >= K;
  int big_guy = max(left_guy, right_guy);
  int bigger_count = bigger_in_range(kill_left, kill_right, orig, big_guy);
  if (!can_use_range) {
    if (bigger_count > 0) {
      return -1;
    }
    return (long long)range * (long long)Y;
  }
  if (range_is_better) {
    int range_times = range / K;
    long long single_times =
        (long long)range - ((long long)range_times * (long long)K);
    return (long long)range_times * (long long)X +
           (long long)single_times * (long long)Y;
  }
  if (bigger_count > 0) {
    int range_times = 1;
    long long single_times =
        (long long)range - ((long long)range_times * (long long)K);
    return (long long)range_times * (long long)X +
           (long long)single_times * (long long)Y;
  }
  return (long long)range * (long long)Y;
}
int main() {
  int N, M;
  cin >> N >> M;
  int X, K, Y;
  cin >> X >> K >> Y;
  map<int, int> elem;
  vector<int> orig(N);
  for (int i = 0; i < N; i++) {
    cin >> orig[i];
    elem[orig[i]] = i;
  }
  vector<int> result(M);
  for (int i = 0; i < M; i++) {
    cin >> result[i];
  }
  if (!canDo(orig, result)) {
    cout << "-1" << endl;
    return 0;
  }
  vector<int> kept_pos;
  for (int i = 0; i < M; i++) {
    kept_pos.push_back(elem[result[i]]);
  }
  long long ans = 0;
  if (kept_pos[0] != 0) {
    long long aux =
        computeAns(0, kept_pos[0] - 1, orig, -1, result[0], X, K, Y);
    if (aux == -1) {
      cout << "-1" << endl;
      return 0;
    }
    ans += aux;
  }
  if (kept_pos[M - 1] != N - 1) {
    long long aux = computeAns(kept_pos[M - 1] + 1, N - 1, orig, result[M - 1],
                               -1, X, K, Y);
    if (aux == -1) {
      cout << "-1" << endl;
      return 0;
    }
    ans += aux;
  }
  if (M > 1) {
    for (int i = 1; i < M; i++) {
      long long aux = computeAns(kept_pos[i - 1] + 1, kept_pos[i] - 1, orig,
                                 result[i - 1], result[i], X, K, Y);
      if (aux == -1) {
        cout << "-1" << endl;
        return 0;
      }
      ans += aux;
    }
  }
  cout << ans << endl;
  return 0;
}

```