#include <bits/stdc++.h>
using namespace std;
void flash() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long x, k, y;
long long dfs(long long i, long long j, long long a[]) {
  if (j - i == 1) return 0;
  long long mx = 0;
  for (long long k = i + 1; k < j; k++) {
    mx = max(mx, a[k]);
  }
  if ((i == -1 or mx > a[i]) and mx > a[j] and j - i - 1 < k) {
    return -1;
  }
  long long c1 = 9000000000000000000, c2 = 9000000000000000000,
            c3 = 9000000000000000000, c4 = 9000000000000000000;
  if ((j - i - 1) % k == 0) {
    c1 = ((j - i - 1) / k) * x;
  } else {
    long long tot = (j - i - 1) / k;
    c4 = tot * x;
    c4 += (j - i - 1 - (tot * k)) * y;
  }
  if (mx < a[i] or mx < a[j]) {
    c2 = (j - i - 1) * y;
  } else {
    c3 = x;
    c3 += (j - i - 1 - k) * y;
  }
  return min({c1, c2, c3, c4});
}
void solve() {
  long long n, m;
  cin >> n >> m;
  cin >> x >> k >> y;
  long long a[n + 10];
  long long b[m + 10];
  long long vis[n + 10];
  long long pos[n + 10];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    vis[a[i]] = 1;
    pos[a[i]] = i;
  }
  long long ok = 1;
  for (long long i = 0; i < m; i++) {
    cin >> b[i];
    if (vis[b[i]] == 0) ok = 0;
  }
  if (m > n or ok == 0) {
    cout << "-1"
         << "\n";
    return;
  }
  long long ans = dfs(-1, pos[b[0]], a);
  long long j = 0;
  b[m] = n + 1;
  pos[n + 1] = n;
  for (long long i = 1; i < m + 1; i++) {
    if (pos[b[i]] < pos[b[i - 1]]) {
      cout << "-1"
           << "\n";
      return;
    }
    long long cost = dfs(pos[b[i - 1]], pos[b[i]], a);
    if (cost == -1) {
      cout << "-1"
           << "\n";
      return;
    }
    ans += cost;
  }
  cout << ans << "\n";
}
int32_t main() {
  flash();
  long long t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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;
void flash() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long x, k, y;
long long dfs(long long i, long long j, long long a[]) {
  if (j - i == 1) return 0;
  long long mx = 0;
  for (long long k = i + 1; k < j; k++) {
    mx = max(mx, a[k]);
  }
  if ((i == -1 or mx > a[i]) and mx > a[j] and j - i - 1 < k) {
    return -1;
  }
  long long c1 = 9000000000000000000, c2 = 9000000000000000000,
            c3 = 9000000000000000000, c4 = 9000000000000000000;
  if ((j - i - 1) % k == 0) {
    c1 = ((j - i - 1) / k) * x;
  } else {
    long long tot = (j - i - 1) / k;
    c4 = tot * x;
    c4 += (j - i - 1 - (tot * k)) * y;
  }
  if (mx < a[i] or mx < a[j]) {
    c2 = (j - i - 1) * y;
  } else {
    c3 = x;
    c3 += (j - i - 1 - k) * y;
  }
  return min({c1, c2, c3, c4});
}
void solve() {
  long long n, m;
  cin >> n >> m;
  cin >> x >> k >> y;
  long long a[n + 10];
  long long b[m + 10];
  long long vis[n + 10];
  long long pos[n + 10];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    vis[a[i]] = 1;
    pos[a[i]] = i;
  }
  long long ok = 1;
  for (long long i = 0; i < m; i++) {
    cin >> b[i];
    if (vis[b[i]] == 0) ok = 0;
  }
  if (m > n or ok == 0) {
    cout << "-1"
         << "\n";
    return;
  }
  long long ans = dfs(-1, pos[b[0]], a);
  long long j = 0;
  b[m] = n + 1;
  pos[n + 1] = n;
  for (long long i = 1; i < m + 1; i++) {
    if (pos[b[i]] < pos[b[i - 1]]) {
      cout << "-1"
           << "\n";
      return;
    }
    long long cost = dfs(pos[b[i - 1]], pos[b[i]], a);
    if (cost == -1) {
      cout << "-1"
           << "\n";
      return;
    }
    ans += cost;
  }
  cout << ans << "\n";
}
int32_t main() {
  flash();
  long long t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```