#include <bits/stdc++.h>
using namespace std;
bool compare(const pair<int long long, int long long> &a,
             const pair<int long long, int long long> &b) {
  return a.second <= b.second;
}
int long long inline power(int long long a, int long long b, int long long p) {
  a %= p;
  int long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % p;
    a = (a * a) % p;
    b >>= 1;
  }
  return ans;
}
int long long inv(int long long n, int long long p) {
  return power(n, p - 2, p);
}
bool inline isPrime(int long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
int long long gcd(int long long a, int long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void solve() {
  int long long n, m, i, j, k, l, r;
  cin >> n >> m;
  int long long x, y;
  cin >> x >> k >> y;
  int long long a[n + 1], b[m + 1];
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (i = 0; i < m; i++) {
    cin >> b[i];
  }
  a[n] = b[m] = 0;
  int long long idx = 0;
  int long long cnt = 0;
  int long long mx = 0;
  int long long pr = -1;
  int long long ans = 0;
  for (i = 0; i < n + 1; i++) {
    if (idx < m + 1 && a[i] == b[idx]) {
      int long long res = 1e18;
      if (mx < a[i] || mx < pr) {
        res = y * cnt;
      }
      for (j = k; j <= cnt; j = j + k) {
        res = min(res, x * (j / k) + y * (cnt - j));
      }
      if (res > 1e17) {
        cout << "-1\n";
        return;
      }
      ans += res;
      idx++;
      pr = a[i];
      mx = 0;
      cnt = 0;
    } else {
      mx = max(mx, a[i]);
      cnt++;
    }
  }
  if (idx != m + 1) {
    cout << "-1\n";
    return;
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int long long t = 1;
  while (t--) solve();
}

### 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;
bool compare(const pair<int long long, int long long> &a,
             const pair<int long long, int long long> &b) {
  return a.second <= b.second;
}
int long long inline power(int long long a, int long long b, int long long p) {
  a %= p;
  int long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % p;
    a = (a * a) % p;
    b >>= 1;
  }
  return ans;
}
int long long inv(int long long n, int long long p) {
  return power(n, p - 2, p);
}
bool inline isPrime(int long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
int long long gcd(int long long a, int long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void solve() {
  int long long n, m, i, j, k, l, r;
  cin >> n >> m;
  int long long x, y;
  cin >> x >> k >> y;
  int long long a[n + 1], b[m + 1];
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (i = 0; i < m; i++) {
    cin >> b[i];
  }
  a[n] = b[m] = 0;
  int long long idx = 0;
  int long long cnt = 0;
  int long long mx = 0;
  int long long pr = -1;
  int long long ans = 0;
  for (i = 0; i < n + 1; i++) {
    if (idx < m + 1 && a[i] == b[idx]) {
      int long long res = 1e18;
      if (mx < a[i] || mx < pr) {
        res = y * cnt;
      }
      for (j = k; j <= cnt; j = j + k) {
        res = min(res, x * (j / k) + y * (cnt - j));
      }
      if (res > 1e17) {
        cout << "-1\n";
        return;
      }
      ans += res;
      idx++;
      pr = a[i];
      mx = 0;
      cnt = 0;
    } else {
      mx = max(mx, a[i]);
      cnt++;
    }
  }
  if (idx != m + 1) {
    cout << "-1\n";
    return;
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int long long t = 1;
  while (t--) solve();
}

```