#include <bits/stdc++.h>
using namespace std;
void OJ() {}
long long n, m, x, k, y;
long long a[200000 + 5], b[200000 + 5];
void imp() { cout << -1 << "\n", exit(0); }
long long remove(long long start, long long end) {
  if (start > end) return 0;
  long long sz = end - start + 1;
  long long maxneb = max(a[start - 1], a[end + 1]);
  long long m = *max_element(a + start, a + end + 1);
  long long a1 = (sz / k) * x + (sz % k) * y;
  if (maxneb > m) return min(a1, y * sz);
  if (sz < k) imp();
  return min(a1, (sz - k) * y + x);
}
void solve() {
  cin >> n >> m >> x >> k >> y;
  for (long long i = 1; i <= n; ++i) cin >> a[i];
  for (long long i = 1; i <= m; ++i) cin >> b[i];
  long long ans = 0;
  long long j = 1;
  for (long long i = 1; i <= m; ++i) {
    long long start = j;
    while (j <= n && a[j] != b[i]) j++;
    if (j > n) imp();
    ans += remove(start, j - 1);
    j++;
  }
  ans += remove(j, n);
  cout << ans << "\n";
}
signed main() {
  OJ();
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}

### 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 OJ() {}
long long n, m, x, k, y;
long long a[200000 + 5], b[200000 + 5];
void imp() { cout << -1 << "\n", exit(0); }
long long remove(long long start, long long end) {
  if (start > end) return 0;
  long long sz = end - start + 1;
  long long maxneb = max(a[start - 1], a[end + 1]);
  long long m = *max_element(a + start, a + end + 1);
  long long a1 = (sz / k) * x + (sz % k) * y;
  if (maxneb > m) return min(a1, y * sz);
  if (sz < k) imp();
  return min(a1, (sz - k) * y + x);
}
void solve() {
  cin >> n >> m >> x >> k >> y;
  for (long long i = 1; i <= n; ++i) cin >> a[i];
  for (long long i = 1; i <= m; ++i) cin >> b[i];
  long long ans = 0;
  long long j = 1;
  for (long long i = 1; i <= m; ++i) {
    long long start = j;
    while (j <= n && a[j] != b[i]) j++;
    if (j > n) imp();
    ans += remove(start, j - 1);
    j++;
  }
  ans += remove(j, n);
  cout << ans << "\n";
}
signed main() {
  OJ();
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}

```