#include <bits/stdc++.h>
using namespace std;
string tostr(long long a) {
  stringstream rr;
  rr << a;
  return rr.str();
}
long long pow(long long c, long long d) { return d == 0 ?: c * pow(c, d - 1); }
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return ((a * b) / gcd(a, b)); }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m, x, y, k;
  cin >> n >> m;
  cin >> x >> k >> y;
  bool flag = false;
  if (m > n) flag = true;
  vector<long long> v(n + 2), r(m + 2), temp;
  v[0] = 0;
  for (long long i = 1; i <= n; i++) cin >> v[i];
  v[n + 1] = 0;
  r[0] = 0;
  for (long long i = 1; i <= m; i++) cin >> r[i];
  r[m + 1] = 0;
  if (flag) {
    cout << -1 << endl;
    return 0;
  }
  long long ans = 0;
  long long ind1, ind2;
  for (ind1 = 0, ind2 = 0; ind1 <= n + 1; ind1++) {
    if (v[ind1] == r[ind2]) {
      ind2++;
      if (temp.size()) {
        sort(temp.begin(), temp.end());
        long long sz = temp.size();
        long long zz = max(r[ind2 - 1], r[ind2 - 2]);
        long long mx = max(temp[0], temp[sz - 1]);
        if (sz < k && mx > zz) {
          cout << -1 << endl;
          return 0;
        }
        if (k * y > x) {
          ans += (sz / k) * x;
          ans += (sz % k) * y;
        } else {
          if (mx > zz) {
            ans += (sz - k) * y;
            ans += x;
          } else
            ans += (sz * y);
        }
      }
      temp.clear();
    } else {
      temp.push_back(v[ind1]);
    }
    if (ind1 + 1 == v.size() && ind2 != r.size()) {
      cout << -1 << endl;
      return 0;
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your task is to create 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;
string tostr(long long a) {
  stringstream rr;
  rr << a;
  return rr.str();
}
long long pow(long long c, long long d) { return d == 0 ?: c * pow(c, d - 1); }
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return ((a * b) / gcd(a, b)); }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m, x, y, k;
  cin >> n >> m;
  cin >> x >> k >> y;
  bool flag = false;
  if (m > n) flag = true;
  vector<long long> v(n + 2), r(m + 2), temp;
  v[0] = 0;
  for (long long i = 1; i <= n; i++) cin >> v[i];
  v[n + 1] = 0;
  r[0] = 0;
  for (long long i = 1; i <= m; i++) cin >> r[i];
  r[m + 1] = 0;
  if (flag) {
    cout << -1 << endl;
    return 0;
  }
  long long ans = 0;
  long long ind1, ind2;
  for (ind1 = 0, ind2 = 0; ind1 <= n + 1; ind1++) {
    if (v[ind1] == r[ind2]) {
      ind2++;
      if (temp.size()) {
        sort(temp.begin(), temp.end());
        long long sz = temp.size();
        long long zz = max(r[ind2 - 1], r[ind2 - 2]);
        long long mx = max(temp[0], temp[sz - 1]);
        if (sz < k && mx > zz) {
          cout << -1 << endl;
          return 0;
        }
        if (k * y > x) {
          ans += (sz / k) * x;
          ans += (sz % k) * y;
        } else {
          if (mx > zz) {
            ans += (sz - k) * y;
            ans += x;
          } else
            ans += (sz * y);
        }
      }
      temp.clear();
    } else {
      temp.push_back(v[ind1]);
    }
    if (ind1 + 1 == v.size() && ind2 != r.size()) {
      cout << -1 << endl;
      return 0;
    }
  }
  cout << ans << endl;
  return 0;
}

```