#include <bits/stdc++.h>
using namespace std;
using pii = pair<long long, long long>;
const long long INF = 1e18;
template <class A, class B>
ostream& operator<<(ostream& os, const pair<A, B> p) {
  return os << "(" << p.first << "," << p.second << ")";
}
void dprint(string s) { cout << "\n"; }
template <class T, class... U>
void dprint(string s, T t, U... u) {
  long long w = s.find(',');
  cout << "[" << s.substr(0, w) << ": " << t << "] ";
  dprint(s.substr(w + 1, (long long)s.size() - w), u...);
}
const long long N = 1e6 + 5;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  long long x, k, y;
  cin >> x >> k >> y;
  long long a[n], b[m];
  for (long long& x : a) cin >> x;
  for (long long& x : b) cin >> x;
  vector<long long> ind = {-1};
  long long cur = 0;
  for (long long i = 0; i < n; i++) {
    if (a[i] == b[cur] and cur < m) {
      ind.push_back(i), cur++;
    }
  }
  ind.push_back(n);
  if (cur != m) return cout << -1, 0;
  auto calc = [&](long long l, long long r, long long mx, long long sz) {
    long long cost = 0;
    long long q = sz / k;
    if (r > mx or l > mx) {
      cost += q * min(x, y * k);
      long long rem = sz % k;
      cost += rem * y;
      return cost;
    } else {
      if (q == 0) {
        cout << -1;
        exit(0);
      } else {
        cost += x;
        q--;
        cost += q * min(x, y * k);
        long long rem = sz % k;
        cost += rem * y;
        return cost;
      }
    }
  };
  long long ans = 0;
  for (long long i = 0; i + 1 < (long long)(ind).size(); i++) {
    bool got = 0;
    long long l, r;
    if (ind[i] == -1)
      l = 0;
    else
      l = a[ind[i]];
    if (ind[i + 1] == n)
      r = 0;
    else
      r = a[ind[i + 1]];
    long long mx = 0, sz = 0;
    ;
    for (long long j = ind[i] + 1; j <= ind[i + 1] - 1; j++) {
      got = 1;
      sz++;
      mx = max(mx, a[j]);
    }
    if (got) {
      long long val = calc(l, r, mx, sz);
      ans += val;
    }
  }
  cout << ans;
  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;
using pii = pair<long long, long long>;
const long long INF = 1e18;
template <class A, class B>
ostream& operator<<(ostream& os, const pair<A, B> p) {
  return os << "(" << p.first << "," << p.second << ")";
}
void dprint(string s) { cout << "\n"; }
template <class T, class... U>
void dprint(string s, T t, U... u) {
  long long w = s.find(',');
  cout << "[" << s.substr(0, w) << ": " << t << "] ";
  dprint(s.substr(w + 1, (long long)s.size() - w), u...);
}
const long long N = 1e6 + 5;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  long long x, k, y;
  cin >> x >> k >> y;
  long long a[n], b[m];
  for (long long& x : a) cin >> x;
  for (long long& x : b) cin >> x;
  vector<long long> ind = {-1};
  long long cur = 0;
  for (long long i = 0; i < n; i++) {
    if (a[i] == b[cur] and cur < m) {
      ind.push_back(i), cur++;
    }
  }
  ind.push_back(n);
  if (cur != m) return cout << -1, 0;
  auto calc = [&](long long l, long long r, long long mx, long long sz) {
    long long cost = 0;
    long long q = sz / k;
    if (r > mx or l > mx) {
      cost += q * min(x, y * k);
      long long rem = sz % k;
      cost += rem * y;
      return cost;
    } else {
      if (q == 0) {
        cout << -1;
        exit(0);
      } else {
        cost += x;
        q--;
        cost += q * min(x, y * k);
        long long rem = sz % k;
        cost += rem * y;
        return cost;
      }
    }
  };
  long long ans = 0;
  for (long long i = 0; i + 1 < (long long)(ind).size(); i++) {
    bool got = 0;
    long long l, r;
    if (ind[i] == -1)
      l = 0;
    else
      l = a[ind[i]];
    if (ind[i + 1] == n)
      r = 0;
    else
      r = a[ind[i + 1]];
    long long mx = 0, sz = 0;
    ;
    for (long long j = ind[i] + 1; j <= ind[i + 1] - 1; j++) {
      got = 1;
      sz++;
      mx = max(mx, a[j]);
    }
    if (got) {
      long long val = calc(l, r, mx, sz);
      ans += val;
    }
  }
  cout << ans;
  return 0;
}

```