#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
ll n;
vector<ll> a, b;
void solve() {
  ll a1, a2, x, y, k, ans = 0;
  cin >> a1 >> a2;
  cin >> x >> k >> y;
  a.push_back(0ll);
  for (ll i = 0; i < a1; i++) {
    ll in;
    cin >> in;
    a.push_back(in);
  }
  a.push_back(0ll);
  b.push_back(0ll);
  for (ll i = 0; i < a2; i++) {
    ll in;
    cin >> in;
    b.push_back(in);
  }
  b.push_back(0ll);
  vector<ll> cal;
  for (ll i = 1, j = 1; i < a.size(); i++) {
    if (a[i] != b[j])
      cal.push_back(a[i]);
    else {
      ll Cals = cal.size();
      if (!Cals) {
        j++;
        continue;
      }
      sort(cal.begin(), cal.end());
      ll Bmax = max(b[j - 1], b[j]), Calmax = cal[Cals - 1];
      if (Calmax > Bmax && Cals < k) {
        cout << "-1" << endl;
        return;
      }
      if (k * y < x) {
        if (Calmax < Bmax)
          ans += Cals * y;
        else
          ans += ((Cals - k) * y) + x;
      } else {
        ll Y = Cals % k;
        ans += ((Cals / k) * x) + (Y * y);
      }
      j++;
      cal.clear();
    }
    if (i + 1 == a.size() && j != b.size()) {
      cout << "-1" << endl;
      return;
    }
  }
  cout << ans << endl;
  return;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  solve();
  return 0;
}

### 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;
using ll = long long;
using ld = long double;
ll n;
vector<ll> a, b;
void solve() {
  ll a1, a2, x, y, k, ans = 0;
  cin >> a1 >> a2;
  cin >> x >> k >> y;
  a.push_back(0ll);
  for (ll i = 0; i < a1; i++) {
    ll in;
    cin >> in;
    a.push_back(in);
  }
  a.push_back(0ll);
  b.push_back(0ll);
  for (ll i = 0; i < a2; i++) {
    ll in;
    cin >> in;
    b.push_back(in);
  }
  b.push_back(0ll);
  vector<ll> cal;
  for (ll i = 1, j = 1; i < a.size(); i++) {
    if (a[i] != b[j])
      cal.push_back(a[i]);
    else {
      ll Cals = cal.size();
      if (!Cals) {
        j++;
        continue;
      }
      sort(cal.begin(), cal.end());
      ll Bmax = max(b[j - 1], b[j]), Calmax = cal[Cals - 1];
      if (Calmax > Bmax && Cals < k) {
        cout << "-1" << endl;
        return;
      }
      if (k * y < x) {
        if (Calmax < Bmax)
          ans += Cals * y;
        else
          ans += ((Cals - k) * y) + x;
      } else {
        ll Y = Cals % k;
        ans += ((Cals / k) * x) + (Y * y);
      }
      j++;
      cal.clear();
    }
    if (i + 1 == a.size() && j != b.size()) {
      cout << "-1" << endl;
      return;
    }
  }
  cout << ans << endl;
  return;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  solve();
  return 0;
}

```