#include <bits/stdc++.h>
using namespace std;
const int N = (int)1e6 + 10;
const int INF = (int)1e9 + 99;
int main() {
  long long n, m, x, y, k;
  cin >> n >> m;
  cin >> x >> k >> y;
  long long a[n + 5], b[m + 5];
  map<long long, long long> pa;
  for (long long i = 1; i <= n; ++i) cin >> a[i], pa[a[i]] = i;
  for (long long i = 1; i <= m; ++i) cin >> b[i];
  b[0] = 0;
  long long ans = 0;
  for (long long i = 1, j = 1; i <= n;) {
    if (j <= m && a[i] == b[j])
      ++i, ++j;
    else {
      long long f = (j <= m) ? pa[b[j]] : n + 1;
      if (f < i) {
        puts("-1");
        return 0;
      }
      long long val = (j <= m) ? max(b[j], b[j - 1]) : b[j - 1];
      long long mx = 0;
      for (long long q = i; q < f; ++q) mx = max(mx, a[q]);
      long long len = f - i;
      if (mx > val && len < k) {
        puts("-1");
        return 0;
      }
      if (len % k) ans += len % k * y, len -= len % k;
      if (mx > val) ans += x, len -= k;
      ans += min(len * y, len / k * x);
      i = f;
    }
    if (i > n && j <= m) {
      puts("-1");
      return 0;
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 int N = (int)1e6 + 10;
const int INF = (int)1e9 + 99;
int main() {
  long long n, m, x, y, k;
  cin >> n >> m;
  cin >> x >> k >> y;
  long long a[n + 5], b[m + 5];
  map<long long, long long> pa;
  for (long long i = 1; i <= n; ++i) cin >> a[i], pa[a[i]] = i;
  for (long long i = 1; i <= m; ++i) cin >> b[i];
  b[0] = 0;
  long long ans = 0;
  for (long long i = 1, j = 1; i <= n;) {
    if (j <= m && a[i] == b[j])
      ++i, ++j;
    else {
      long long f = (j <= m) ? pa[b[j]] : n + 1;
      if (f < i) {
        puts("-1");
        return 0;
      }
      long long val = (j <= m) ? max(b[j], b[j - 1]) : b[j - 1];
      long long mx = 0;
      for (long long q = i; q < f; ++q) mx = max(mx, a[q]);
      long long len = f - i;
      if (mx > val && len < k) {
        puts("-1");
        return 0;
      }
      if (len % k) ans += len % k * y, len -= len % k;
      if (mx > val) ans += x, len -= k;
      ans += min(len * y, len / k * x);
      i = f;
    }
    if (i > n && j <= m) {
      puts("-1");
      return 0;
    }
  }
  cout << ans << endl;
  return 0;
}

```