#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long tc = 1;
  while (tc--) {
    long long n, m, x, k, y;
    cin >> n >> m >> x >> k >> y;
    vector<long long> a(n), b(m), v(1, -1);
    for (long long i = 0; i < (n); i++) cin >> a[i];
    for (long long i = 0; i < (m); i++) cin >> b[i];
    if (m > n) {
      cout << -1 << "\n";
      continue;
    }
    bool flag = 0;
    for (long long i = 0, j = 0; i < m; i++, j++) {
      while (j < n && a[j] != b[i]) {
        j++;
      }
      if (j == n) {
        flag = 1;
        break;
      }
      v.push_back(j);
    }
    if (flag) {
      cout << -1 << "\n";
      continue;
    }
    v.push_back(n);
    long long ans = 0;
    if (x <= y * k) flag = 1;
    for (long long i = 0; i < (v.size() - 1); i++) {
      long long len = v[i + 1] - v[i] - 1;
      long long segmi = max((i > 0 ? a[v[i]] : a[v[i + 1]]),
                            (i < v.size() - 2 ? a[v[i + 1]] : a[v[i]])),
                segm = segmi;
      for (long long j = v[i] + 1; j < v[i + 1]; j++) {
        segm = max(segm, a[j]);
      }
      if (segm > segmi && len < k) {
        ans = -1;
        break;
      } else if (segm > segmi) {
        if (flag)
          ans += (len / k) * x + (len % k) * y;
        else
          ans += x + (len - k) * y;
      } else {
        if (flag)
          ans += (len / k) * x + (len % k) * y;
        else
          ans += len * y;
      }
    }
    cout << ans << "\n";
  }
  cerr << "Time : " << ((double)clock()) / (double)CLOCKS_PER_SEC << "s\n";
  return 0;
}

### 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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long tc = 1;
  while (tc--) {
    long long n, m, x, k, y;
    cin >> n >> m >> x >> k >> y;
    vector<long long> a(n), b(m), v(1, -1);
    for (long long i = 0; i < (n); i++) cin >> a[i];
    for (long long i = 0; i < (m); i++) cin >> b[i];
    if (m > n) {
      cout << -1 << "\n";
      continue;
    }
    bool flag = 0;
    for (long long i = 0, j = 0; i < m; i++, j++) {
      while (j < n && a[j] != b[i]) {
        j++;
      }
      if (j == n) {
        flag = 1;
        break;
      }
      v.push_back(j);
    }
    if (flag) {
      cout << -1 << "\n";
      continue;
    }
    v.push_back(n);
    long long ans = 0;
    if (x <= y * k) flag = 1;
    for (long long i = 0; i < (v.size() - 1); i++) {
      long long len = v[i + 1] - v[i] - 1;
      long long segmi = max((i > 0 ? a[v[i]] : a[v[i + 1]]),
                            (i < v.size() - 2 ? a[v[i + 1]] : a[v[i]])),
                segm = segmi;
      for (long long j = v[i] + 1; j < v[i + 1]; j++) {
        segm = max(segm, a[j]);
      }
      if (segm > segmi && len < k) {
        ans = -1;
        break;
      } else if (segm > segmi) {
        if (flag)
          ans += (len / k) * x + (len % k) * y;
        else
          ans += x + (len - k) * y;
      } else {
        if (flag)
          ans += (len / k) * x + (len % k) * y;
        else
          ans += len * y;
      }
    }
    cout << ans << "\n";
  }
  cerr << "Time : " << ((double)clock()) / (double)CLOCKS_PER_SEC << "s\n";
  return 0;
}

```