#include <bits/stdc++.h>
using ll = long long;
using ld = long double;
using ull = unsigned long long;
using namespace std;
const int N = 2e5 + 3;
int n, m, x, k, y, a[N], b[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> m >> x >> k >> y;
  for (int i = (1); i <= (n); ++i) cin >> a[i];
  for (int i = (1); i <= (m); ++i) cin >> b[i];
  b[m + 1] = a[n + 1] = 0;
  n++, m++;
  if (m > n) {
    cout << -1;
    return 0;
  }
  int ib = 1, ia = 1;
  ll ans = 0;
  bool ok = 1;
  while (ib <= m && ia <= n) {
    int mx = 0, lia = ia;
    while (ia <= n && a[ia] != b[ib]) {
      mx = max(mx, a[ia]);
      ia++;
    }
    if (ia > n) break;
    if (mx > max(b[ib], b[ib - 1]) && (ia - lia) < k) break;
    int num = ia - lia;
    if (k * 1ll * y >= x)
      ans += (num / k) * 1ll * x + (num % k) * 1ll * y;
    else {
      if (mx > max(b[ib], b[ib - 1]))
        ans += x + (num - k) * 1ll * y;
      else
        ans += num * 1ll * y;
    }
    ia++, ib++;
  }
  if (ib <= m || ia <= n) ok = 0;
  if (ok)
    cout << ans;
  else
    cout << -1;
  return 0;
}

### Prompt

Your challenge is to write 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 ll = long long;
using ld = long double;
using ull = unsigned long long;
using namespace std;
const int N = 2e5 + 3;
int n, m, x, k, y, a[N], b[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> m >> x >> k >> y;
  for (int i = (1); i <= (n); ++i) cin >> a[i];
  for (int i = (1); i <= (m); ++i) cin >> b[i];
  b[m + 1] = a[n + 1] = 0;
  n++, m++;
  if (m > n) {
    cout << -1;
    return 0;
  }
  int ib = 1, ia = 1;
  ll ans = 0;
  bool ok = 1;
  while (ib <= m && ia <= n) {
    int mx = 0, lia = ia;
    while (ia <= n && a[ia] != b[ib]) {
      mx = max(mx, a[ia]);
      ia++;
    }
    if (ia > n) break;
    if (mx > max(b[ib], b[ib - 1]) && (ia - lia) < k) break;
    int num = ia - lia;
    if (k * 1ll * y >= x)
      ans += (num / k) * 1ll * x + (num % k) * 1ll * y;
    else {
      if (mx > max(b[ib], b[ib - 1]))
        ans += x + (num - k) * 1ll * y;
      else
        ans += num * 1ll * y;
    }
    ia++, ib++;
  }
  if (ib <= m || ia <= n) ok = 0;
  if (ok)
    cout << ans;
  else
    cout << -1;
  return 0;
}

```