#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5, M = 1;
int n, m, a[N], vis[N];
long long x, k, y;
void solve() {
  cin >> n >> m;
  cin >> x >> k >> y;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int pos = 0, f = 0;
  for (int i = 1; i <= m; i++) {
    int t;
    scanf("%d", &t);
    while (++pos <= n) {
      if (a[pos] == t) {
        vis[pos] = 1;
        break;
      }
    }
    if (pos == n + 1) {
      f = 1;
    }
  }
  if (f) {
    puts("-1");
    return;
  }
  vector<int> v;
  int check = 0;
  if (x < y * k)
    check = 1;
  else if (x > y * k)
    check = 2;
  else
    check = 3;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    if (!vis[i])
      v.push_back(a[i]);
    else {
      if (v.empty()) continue;
      int mx = *max_element(v.begin(), v.end());
      if (a[i] > mx || a[i - v.size() - 1] > mx) {
        if (check == 3 || check == 2 || v.size() < k)
          ans += y * v.size();
        else if (check == 1)
          ans += v.size() / k * x + (v.size() - v.size() / k * k) * y;
      } else {
        if (v.size() < k) {
          puts("-1");
          return;
        }
        if (check == 3 || check == 2) {
          ans += x + (v.size() - k) * y;
        } else if (check == 1)
          ans += v.size() / k * x + (v.size() - v.size() / k * k) * y;
      }
      v.clear();
    }
  }
  if (!v.empty()) {
    int mx = *max_element(v.begin(), v.end());
    if (a[n - v.size()] > mx) {
      if (check == 3 || check == 2 || v.size() < k)
        ans += y * v.size();
      else if (check == 1)
        ans += v.size() / k * x + (v.size() - v.size() / k * k) * y;
    } else {
      if (v.size() < k) {
        puts("-1");
        return;
      }
      if (check == 3 || check == 2) {
        ans += x + (v.size() - k) * y;
      } else if (check == 1)
        ans += v.size() / k * x + (v.size() - v.size() / k * k) * y;
    }
  }
  cout << ans << '\n';
}
int main() {
  solve();
  return 0;
}

### Prompt

Your challenge is to write 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 int N = 2e5 + 5, M = 1;
int n, m, a[N], vis[N];
long long x, k, y;
void solve() {
  cin >> n >> m;
  cin >> x >> k >> y;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int pos = 0, f = 0;
  for (int i = 1; i <= m; i++) {
    int t;
    scanf("%d", &t);
    while (++pos <= n) {
      if (a[pos] == t) {
        vis[pos] = 1;
        break;
      }
    }
    if (pos == n + 1) {
      f = 1;
    }
  }
  if (f) {
    puts("-1");
    return;
  }
  vector<int> v;
  int check = 0;
  if (x < y * k)
    check = 1;
  else if (x > y * k)
    check = 2;
  else
    check = 3;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    if (!vis[i])
      v.push_back(a[i]);
    else {
      if (v.empty()) continue;
      int mx = *max_element(v.begin(), v.end());
      if (a[i] > mx || a[i - v.size() - 1] > mx) {
        if (check == 3 || check == 2 || v.size() < k)
          ans += y * v.size();
        else if (check == 1)
          ans += v.size() / k * x + (v.size() - v.size() / k * k) * y;
      } else {
        if (v.size() < k) {
          puts("-1");
          return;
        }
        if (check == 3 || check == 2) {
          ans += x + (v.size() - k) * y;
        } else if (check == 1)
          ans += v.size() / k * x + (v.size() - v.size() / k * k) * y;
      }
      v.clear();
    }
  }
  if (!v.empty()) {
    int mx = *max_element(v.begin(), v.end());
    if (a[n - v.size()] > mx) {
      if (check == 3 || check == 2 || v.size() < k)
        ans += y * v.size();
      else if (check == 1)
        ans += v.size() / k * x + (v.size() - v.size() / k * k) * y;
    } else {
      if (v.size() < k) {
        puts("-1");
        return;
      }
      if (check == 3 || check == 2) {
        ans += x + (v.size() - k) * y;
      } else if (check == 1)
        ans += v.size() / k * x + (v.size() - v.size() / k * k) * y;
    }
  }
  cout << ans << '\n';
}
int main() {
  solve();
  return 0;
}

```