#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e9 + 7;
const long long N = 2e5;
long long a[N], b[N];
long long n, m, x, y, k;
long long solve(long long l, long long r, long long mx) {
  if (l > r) return 0;
  vector<long long> v;
  for (long long i = l; i <= r; ++i) v.push_back(a[i]);
  long long p = 0;
  for (long long u : v) p = max(p, u);
  if (v.size() < k) {
    if (p > mx)
      return -1;
    else
      return y * v.size();
  }
  long long ans = 0;
  ans += (v.size() % k) * y;
  long long sz = v.size() - v.size() % k;
  if (y * k < x) {
    if (p < mx)
      return ans + sz * y;
    else
      return ans + (sz - k) * y + x;
  } else {
    return ans + sz / k * x;
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  ;
  cin >> n >> m >> x >> k >> y;
  for (long long i = 0; i < n; ++i) cin >> a[i];
  for (long long i = 0; i < m; ++i) cin >> b[i];
  vector<long long> ps;
  long long pt = 0;
  for (long long i = 0; i < n; ++i) {
    if (pt == m) break;
    if (b[pt] == a[i]) {
      ps.push_back(i);
      ++pt;
    }
  }
  if (pt != m) {
    cout << -1;
    return 0;
  }
  long long ans = 0;
  for (long long i = 0; i < ps.size(); ++i) {
    if (!i) {
      long long sl = solve(0, ps[i] - 1, a[ps[i]]);
      if (sl == -1) {
        cout << -1;
        return 0;
      }
      ans += sl;
    } else {
      long long sl =
          solve(ps[i - 1] + 1, ps[i] - 1, max(a[ps[i]], a[ps[i - 1]]));
      if (sl == -1) {
        cout << -1;
        return 0;
      }
      ans += sl;
    }
  }
  long long sl = solve(ps[ps.size() - 1] + 1, n - 1, a[ps[ps.size() - 1]]);
  if (sl == -1) {
    cout << sl;
    return 0;
  }
  ans += sl;
  cout << ans;
}

### Prompt

Create a solution in Cpp for 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;
const long long inf = 1e9 + 7;
const long long N = 2e5;
long long a[N], b[N];
long long n, m, x, y, k;
long long solve(long long l, long long r, long long mx) {
  if (l > r) return 0;
  vector<long long> v;
  for (long long i = l; i <= r; ++i) v.push_back(a[i]);
  long long p = 0;
  for (long long u : v) p = max(p, u);
  if (v.size() < k) {
    if (p > mx)
      return -1;
    else
      return y * v.size();
  }
  long long ans = 0;
  ans += (v.size() % k) * y;
  long long sz = v.size() - v.size() % k;
  if (y * k < x) {
    if (p < mx)
      return ans + sz * y;
    else
      return ans + (sz - k) * y + x;
  } else {
    return ans + sz / k * x;
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  ;
  cin >> n >> m >> x >> k >> y;
  for (long long i = 0; i < n; ++i) cin >> a[i];
  for (long long i = 0; i < m; ++i) cin >> b[i];
  vector<long long> ps;
  long long pt = 0;
  for (long long i = 0; i < n; ++i) {
    if (pt == m) break;
    if (b[pt] == a[i]) {
      ps.push_back(i);
      ++pt;
    }
  }
  if (pt != m) {
    cout << -1;
    return 0;
  }
  long long ans = 0;
  for (long long i = 0; i < ps.size(); ++i) {
    if (!i) {
      long long sl = solve(0, ps[i] - 1, a[ps[i]]);
      if (sl == -1) {
        cout << -1;
        return 0;
      }
      ans += sl;
    } else {
      long long sl =
          solve(ps[i - 1] + 1, ps[i] - 1, max(a[ps[i]], a[ps[i - 1]]));
      if (sl == -1) {
        cout << -1;
        return 0;
      }
      ans += sl;
    }
  }
  long long sl = solve(ps[ps.size() - 1] + 1, n - 1, a[ps[ps.size() - 1]]);
  if (sl == -1) {
    cout << sl;
    return 0;
  }
  ans += sl;
  cout << ans;
}

```