#include <bits/stdc++.h>
using namespace std;
const long long M = 1e9 + 7;
void solve() {
  long long n, m;
  cin >> n >> m;
  long long x, k, y;
  cin >> x >> k >> y;
  long long a[n + 2];
  memset(a, 0, sizeof(a));
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  long long b[m + 2];
  memset(b, 0, sizeof(b));
  for (long long i = 1; i <= m; i++) {
    cin >> b[i];
  }
  long long cur = 0;
  vector<pair<long long, long long>> vp;
  pair<long long, long long> temp = {0, -1};
  for (long long i = 0; i <= n + 1; i++) {
    if (cur < m + 2 && b[cur] == a[i]) {
      temp.second = i - 1;
      if (temp.first > 0 && temp.second > 0 && temp.first < n + 1 &&
          temp.second < n + 1 && temp.second >= temp.first) {
        vp.push_back(temp);
      }
      temp = {0, -1};
      cur++;
    } else {
      if (temp.first == 0) temp.first = i;
    }
  }
  if (cur < m + 2) {
    cout << -1 << "\n";
    return;
  }
  long long ans = 0;
  while (!vp.empty()) {
    auto it = vp.back();
    long long start_index = it.first;
    long long end_index = it.second;
    long long left = a[start_index - 1];
    long long right = a[end_index + 1];
    vp.pop_back();
    long long mx = 0;
    for (long long i = start_index; i <= end_index; i++) {
      mx = max(mx, a[i]);
    }
    long long total = end_index - start_index + 1;
    if (k * y > x) {
      if (left > mx || right > mx) {
        ans += (total / k) * x + (total % k) * y;
      } else {
        if (total >= k) {
          ans += (total / k) * x + (total % k) * y;
        } else {
          cout << -1 << "\n";
          return;
        }
      }
    } else {
      if (left > mx || right > mx) {
        ans += total * y;
      } else {
        if (total >= k) {
          ans += x + (total - k) * y;
        } else {
          cout << -1 << "\n";
          return;
        }
      }
    }
  }
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t = 1;
  while (t--) {
    solve();
  }
}

### 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;
const long long M = 1e9 + 7;
void solve() {
  long long n, m;
  cin >> n >> m;
  long long x, k, y;
  cin >> x >> k >> y;
  long long a[n + 2];
  memset(a, 0, sizeof(a));
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  long long b[m + 2];
  memset(b, 0, sizeof(b));
  for (long long i = 1; i <= m; i++) {
    cin >> b[i];
  }
  long long cur = 0;
  vector<pair<long long, long long>> vp;
  pair<long long, long long> temp = {0, -1};
  for (long long i = 0; i <= n + 1; i++) {
    if (cur < m + 2 && b[cur] == a[i]) {
      temp.second = i - 1;
      if (temp.first > 0 && temp.second > 0 && temp.first < n + 1 &&
          temp.second < n + 1 && temp.second >= temp.first) {
        vp.push_back(temp);
      }
      temp = {0, -1};
      cur++;
    } else {
      if (temp.first == 0) temp.first = i;
    }
  }
  if (cur < m + 2) {
    cout << -1 << "\n";
    return;
  }
  long long ans = 0;
  while (!vp.empty()) {
    auto it = vp.back();
    long long start_index = it.first;
    long long end_index = it.second;
    long long left = a[start_index - 1];
    long long right = a[end_index + 1];
    vp.pop_back();
    long long mx = 0;
    for (long long i = start_index; i <= end_index; i++) {
      mx = max(mx, a[i]);
    }
    long long total = end_index - start_index + 1;
    if (k * y > x) {
      if (left > mx || right > mx) {
        ans += (total / k) * x + (total % k) * y;
      } else {
        if (total >= k) {
          ans += (total / k) * x + (total % k) * y;
        } else {
          cout << -1 << "\n";
          return;
        }
      }
    } else {
      if (left > mx || right > mx) {
        ans += total * y;
      } else {
        if (total >= k) {
          ans += x + (total - k) * y;
        } else {
          cout << -1 << "\n";
          return;
        }
      }
    }
  }
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t = 1;
  while (t--) {
    solve();
  }
}

```