#include <bits/stdc++.h>
using namespace std;
signed 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 i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < m; i++) cin >> b[i];
  long long p = 0;
  vector<long long> v;
  v.push_back(-1);
  for (long long i = 0; i < n; i++) {
    if (p < m && a[i] == b[p]) {
      v.push_back(i);
      p++;
    }
  }
  v.push_back(n);
  if (p != m) {
    cout << -1 << endl;
    return 0;
  }
  bool ok = 1;
  long long ans = 0;
  for (long long i = 1; i < v.size(); i++) {
    long long l = v[i - 1], r = v[i];
    long long len = r - l - 1;
    long long m = -5;
    for (long long i = l + 1; i < r; i++) {
      m = max(m, a[i]);
    }
    long long v1 = -50, v2 = -50;
    if (l != -1) v1 = a[l];
    if (r != n) v2 = a[r];
    long long curr = 0;
    if (len % k != 0) {
      if (len < k and v1 < m and v2 < m) {
        ok = 0;
        break;
      } else {
        curr += y * (len % k);
      }
    }
    len -= len % k;
    if (len == 0) {
      ans += curr;
      continue;
    }
    long long val = x * (len / k);
    if (v1 > m or v2 > m) {
      val = min(val, y * (len));
    } else {
      val = min(val, x + y * (len - k));
    }
    curr += val;
    ans += curr;
  }
  if (ok)
    cout << ans << endl;
  else
    cout << -1 << endl;
}

### 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;
signed 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 i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < m; i++) cin >> b[i];
  long long p = 0;
  vector<long long> v;
  v.push_back(-1);
  for (long long i = 0; i < n; i++) {
    if (p < m && a[i] == b[p]) {
      v.push_back(i);
      p++;
    }
  }
  v.push_back(n);
  if (p != m) {
    cout << -1 << endl;
    return 0;
  }
  bool ok = 1;
  long long ans = 0;
  for (long long i = 1; i < v.size(); i++) {
    long long l = v[i - 1], r = v[i];
    long long len = r - l - 1;
    long long m = -5;
    for (long long i = l + 1; i < r; i++) {
      m = max(m, a[i]);
    }
    long long v1 = -50, v2 = -50;
    if (l != -1) v1 = a[l];
    if (r != n) v2 = a[r];
    long long curr = 0;
    if (len % k != 0) {
      if (len < k and v1 < m and v2 < m) {
        ok = 0;
        break;
      } else {
        curr += y * (len % k);
      }
    }
    len -= len % k;
    if (len == 0) {
      ans += curr;
      continue;
    }
    long long val = x * (len / k);
    if (v1 > m or v2 > m) {
      val = min(val, y * (len));
    } else {
      val = min(val, x + y * (len - k));
    }
    curr += val;
    ans += curr;
  }
  if (ok)
    cout << ans << endl;
  else
    cout << -1 << endl;
}

```