#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m, x, k, y;
  cin >> n >> m;
  cin >> x >> k >> y;
  vector<int> a(n);
  for (int i = (0); i < (n); ++i) cin >> a[i];
  vector<int> b(n);
  for (int i = (0); i < (m); ++i) cin >> b[i];
  int bdex = 0;
  vector<pair<int, int>> borders;
  int lef = -1;
  int gcnt = 0;
  for (int i = (0); i < (n); ++i) {
    if (bdex < m && b[bdex] == a[i]) {
      if (gcnt > 0) {
        borders.push_back({lef, i});
      }
      lef = i;
      ++bdex;
      gcnt = 0;
    } else {
      ++gcnt;
      if (i == n - 1) borders.push_back({lef, i + 1});
    }
  }
  if (bdex < m) {
    cout << -1 << '\n';
    return 0;
  }
  bool fireok = x <= y * k;
  long long ans = 0;
  int mx;
  bool mxatbor;
  for (auto& bor : borders) {
    mx = *max_element(begin(a) + max(0, bor.first),
                      begin(a) + min(bor.second + 1, n));
    mxatbor = (bor.first >= 0 && mx == a[bor.first]) ||
              (bor.second < n && mx == a[bor.second]);
    if (!mxatbor && bor.second - bor.first - 1 < k) {
      cout << -1 << '\n';
      return 0;
    }
    if (!mxatbor) {
      ans += x;
      bor.first += k;
    }
    if (fireok) {
      while (bor.second - bor.first - 1 >= k) {
        ans += x;
        bor.first += k;
      }
    }
    ans += (long long)(bor.second - bor.first - 1) * (long long)y;
  }
  cout << ans << '\n';
  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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m, x, k, y;
  cin >> n >> m;
  cin >> x >> k >> y;
  vector<int> a(n);
  for (int i = (0); i < (n); ++i) cin >> a[i];
  vector<int> b(n);
  for (int i = (0); i < (m); ++i) cin >> b[i];
  int bdex = 0;
  vector<pair<int, int>> borders;
  int lef = -1;
  int gcnt = 0;
  for (int i = (0); i < (n); ++i) {
    if (bdex < m && b[bdex] == a[i]) {
      if (gcnt > 0) {
        borders.push_back({lef, i});
      }
      lef = i;
      ++bdex;
      gcnt = 0;
    } else {
      ++gcnt;
      if (i == n - 1) borders.push_back({lef, i + 1});
    }
  }
  if (bdex < m) {
    cout << -1 << '\n';
    return 0;
  }
  bool fireok = x <= y * k;
  long long ans = 0;
  int mx;
  bool mxatbor;
  for (auto& bor : borders) {
    mx = *max_element(begin(a) + max(0, bor.first),
                      begin(a) + min(bor.second + 1, n));
    mxatbor = (bor.first >= 0 && mx == a[bor.first]) ||
              (bor.second < n && mx == a[bor.second]);
    if (!mxatbor && bor.second - bor.first - 1 < k) {
      cout << -1 << '\n';
      return 0;
    }
    if (!mxatbor) {
      ans += x;
      bor.first += k;
    }
    if (fireok) {
      while (bor.second - bor.first - 1 >= k) {
        ans += x;
        bor.first += k;
      }
    }
    ans += (long long)(bor.second - bor.first - 1) * (long long)y;
  }
  cout << ans << '\n';
  return 0;
}

```