#include <bits/stdc++.h>
using namespace std;
long long v[200005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n, m, i, j, t, x, y, k;
  cin >> n >> m >> x >> k >> y;
  for (i = 1; i <= n; ++i) cin >> v[i];
  i = 1;
  long long rz = 0;
  long long ap = 0;
  for (j = 1; j <= m; ++j) {
    long long a;
    cin >> a;
    long long cnt = 0;
    long long mxc = 0;
    while (i <= n && v[i] != a) {
      mxc = max(mxc, v[i]);
      ++i;
      ++cnt;
    }
    if (i == n + 1) {
      cout << -1;
      return 0;
    }
    if (mxc < max(a, ap)) {
      rz += y * (cnt % k);
      cnt -= cnt % k;
      rz += min(cnt * 1LL * y, x * 1LL * cnt / k);
    } else {
      if (cnt == cnt % k) {
        cout << -1;
        return 0;
      }
      rz += y * (cnt % k);
      cnt -= cnt % k;
      rz += x;
      cnt -= k;
      rz += min(cnt * y, x * cnt / k);
    }
    ++i;
    ap = a;
  }
  long long mxc = 0;
  long long a = 0;
  long long cnt = 0;
  while (i <= n && v[i] != a) {
    mxc = max(mxc, v[i]);
    ++i;
    ++cnt;
  }
  if (mxc < max(a, ap)) {
    rz += y * (cnt % k);
    cnt -= cnt % k;
    rz += min(cnt * 1LL * y, x * 1LL * cnt / k);
  } else {
    if (cnt == cnt % k) {
      cout << -1;
      return 0;
    }
    rz += y * (cnt % k);
    cnt -= cnt % k;
    rz += x;
    cnt -= k;
    rz += min(cnt * y, x * cnt / k);
  }
  cout << rz;
  return 0;
}

### 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;
long long v[200005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n, m, i, j, t, x, y, k;
  cin >> n >> m >> x >> k >> y;
  for (i = 1; i <= n; ++i) cin >> v[i];
  i = 1;
  long long rz = 0;
  long long ap = 0;
  for (j = 1; j <= m; ++j) {
    long long a;
    cin >> a;
    long long cnt = 0;
    long long mxc = 0;
    while (i <= n && v[i] != a) {
      mxc = max(mxc, v[i]);
      ++i;
      ++cnt;
    }
    if (i == n + 1) {
      cout << -1;
      return 0;
    }
    if (mxc < max(a, ap)) {
      rz += y * (cnt % k);
      cnt -= cnt % k;
      rz += min(cnt * 1LL * y, x * 1LL * cnt / k);
    } else {
      if (cnt == cnt % k) {
        cout << -1;
        return 0;
      }
      rz += y * (cnt % k);
      cnt -= cnt % k;
      rz += x;
      cnt -= k;
      rz += min(cnt * y, x * cnt / k);
    }
    ++i;
    ap = a;
  }
  long long mxc = 0;
  long long a = 0;
  long long cnt = 0;
  while (i <= n && v[i] != a) {
    mxc = max(mxc, v[i]);
    ++i;
    ++cnt;
  }
  if (mxc < max(a, ap)) {
    rz += y * (cnt % k);
    cnt -= cnt % k;
    rz += min(cnt * 1LL * y, x * 1LL * cnt / k);
  } else {
    if (cnt == cnt % k) {
      cout << -1;
      return 0;
    }
    rz += y * (cnt % k);
    cnt -= cnt % k;
    rz += x;
    cnt -= k;
    rz += min(cnt * y, x * cnt / k);
  }
  cout << rz;
  return 0;
}

```