#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
using namespace std;
const int N = 200003, mod = 1e9 + 7;
const long long inf = (1e18 + 11);
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long ans = 0, di, ma, k;
  int n, m;
  cin >> n >> m;
  int x, y;
  cin >> x >> k >> y;
  int a[n + 3], b[m + 3], i, j, l;
  a[0] = 0, a[n + 1] = 0, b[0] = 0, b[m + 1] = 0;
  for (i = 1; i <= n; ++i) cin >> a[i];
  for (i = 1; i <= m; ++i) cin >> b[i];
  for (i = 1, j = 1; i <= n; ++i) {
    if (j > m) goto takecare;
    if (a[i] == b[j]) {
      ++j;
      continue;
    } else {
    takecare:
      l = i, ma = 0;
      while (i <= n and a[i] != b[j]) ma = max(ma, a[i] * 1ll), ++i;
      if (i > n and j <= m) return cout << -1, 0;
      di = i - l;
      if (ma < b[j] or ma < b[j - 1]) {
        if (x <= k * y)
          ans += (di / k) * x + (di % k) * y;
        else
          ans += di * y;
      } else {
        if (di < k)
          return cout << -1, 0;
        else {
          if (x <= k * y)
            ans += (di / k) * x + (di % k) * y;
          else
            ans += x + (di - k) * y;
        }
      }
      if (j > m) break;
      ++j;
    }
  }
  if (j <= m)
    cout << -1;
  else
    cout << ans;
  return 0;
}

### Prompt

Please formulate 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>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
using namespace std;
const int N = 200003, mod = 1e9 + 7;
const long long inf = (1e18 + 11);
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long ans = 0, di, ma, k;
  int n, m;
  cin >> n >> m;
  int x, y;
  cin >> x >> k >> y;
  int a[n + 3], b[m + 3], i, j, l;
  a[0] = 0, a[n + 1] = 0, b[0] = 0, b[m + 1] = 0;
  for (i = 1; i <= n; ++i) cin >> a[i];
  for (i = 1; i <= m; ++i) cin >> b[i];
  for (i = 1, j = 1; i <= n; ++i) {
    if (j > m) goto takecare;
    if (a[i] == b[j]) {
      ++j;
      continue;
    } else {
    takecare:
      l = i, ma = 0;
      while (i <= n and a[i] != b[j]) ma = max(ma, a[i] * 1ll), ++i;
      if (i > n and j <= m) return cout << -1, 0;
      di = i - l;
      if (ma < b[j] or ma < b[j - 1]) {
        if (x <= k * y)
          ans += (di / k) * x + (di % k) * y;
        else
          ans += di * y;
      } else {
        if (di < k)
          return cout << -1, 0;
        else {
          if (x <= k * y)
            ans += (di / k) * x + (di % k) * y;
          else
            ans += x + (di - k) * y;
        }
      }
      if (j > m) break;
      ++j;
    }
  }
  if (j <= m)
    cout << -1;
  else
    cout << ans;
  return 0;
}

```