#include <bits/stdc++.h>
using namespace std;
long long a[(200010)], b[(200010)], del[(200010)];
vector<long long> L;
signed main() {
  {
    long long n, m;
    cin >> n >> m;
    long long x, k, y;
    cin >> x >> k >> y;
    for (long long i = 1; i <= n; i++) {
      scanf("%lld", &a[i]);
      del[i] = 0;
    }
    for (long long i = 1; i <= m; i++) scanf("%lld", &b[i]);
    long long i = 1, j = 1, N = 0;
    for (; i <= n; i++) {
      if (a[i] == b[j])
        j++;
      else {
        del[i] = 1;
        N++;
      }
    }
    if (j <= m) {
      printf("-1\n");
      return 0;
    }
    long long M = 0;
    long long mx = 0, left = 0;
    a[n + 1] = 0;
    bool ok = true;
    for (long long i = 1; i <= n + 1; i++) {
      if (del[i]) {
        mx = a[i] > a[mx] ? i : mx;
      } else {
        if (!mx)
          left = i;
        else {
          L.push_back(i - left - 1);
          if (a[mx] > a[left] && a[mx] > a[i]) {
            M++;
            if (i - left - 1 < k) {
              ok = false;
              break;
            }
          }
          mx = 0;
          left = i;
        }
      }
    }
    if (!ok) {
      printf("-1\n");
      return 0;
    }
    long long ans = 0;
    if (x >= k * y) {
      ans = M * x + (N - M * k) * y;
    } else {
      for (long long i = 0, sz = L.size(); i < sz; i++) {
        ans += (L[i] / k) * x + (L[i] % k) * y;
      }
    }
    cout << ans << endl;
  }
  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;
long long a[(200010)], b[(200010)], del[(200010)];
vector<long long> L;
signed main() {
  {
    long long n, m;
    cin >> n >> m;
    long long x, k, y;
    cin >> x >> k >> y;
    for (long long i = 1; i <= n; i++) {
      scanf("%lld", &a[i]);
      del[i] = 0;
    }
    for (long long i = 1; i <= m; i++) scanf("%lld", &b[i]);
    long long i = 1, j = 1, N = 0;
    for (; i <= n; i++) {
      if (a[i] == b[j])
        j++;
      else {
        del[i] = 1;
        N++;
      }
    }
    if (j <= m) {
      printf("-1\n");
      return 0;
    }
    long long M = 0;
    long long mx = 0, left = 0;
    a[n + 1] = 0;
    bool ok = true;
    for (long long i = 1; i <= n + 1; i++) {
      if (del[i]) {
        mx = a[i] > a[mx] ? i : mx;
      } else {
        if (!mx)
          left = i;
        else {
          L.push_back(i - left - 1);
          if (a[mx] > a[left] && a[mx] > a[i]) {
            M++;
            if (i - left - 1 < k) {
              ok = false;
              break;
            }
          }
          mx = 0;
          left = i;
        }
      }
    }
    if (!ok) {
      printf("-1\n");
      return 0;
    }
    long long ans = 0;
    if (x >= k * y) {
      ans = M * x + (N - M * k) * y;
    } else {
      for (long long i = 0, sz = L.size(); i < sz; i++) {
        ans += (L[i] / k) * x + (L[i] % k) * y;
      }
    }
    cout << ans << endl;
  }
  return 0;
}

```