#include <bits/stdc++.h>
const long long INFLL = ~((long long)1 << 63);
const int N = 200000;
int n, m, x, y, k, a[N + 1];
int main() {
  scanf("%d%d%d%d%d", &n, &m, &x, &k, &y);
  for (int i = 0; i < n; ++i) scanf("%d", a + i);
  int pr = -1, v;
  bool ok = true;
  long long ans = 0;
  for (int i = 0, j = 0; j <= m && ok; ++i, ++j) {
    if (j < m)
      scanf("%d", &v);
    else
      v = 0;
    int f = i, mx = 0;
    while (i <= n && a[i] != v) {
      mx = std::max(mx, a[i]);
      ++i;
    }
    if (i == n + 1)
      ok = false;
    else {
      long long cur = INFLL;
      if (i - f >= k)
        cur = std::min(
            (long long)x * ((i - f) / k) + (long long)y * ((i - f) % k),
            x + (long long)y * (i - f - k));
      if (mx < std::max(pr, v)) cur = std::min(cur, (long long)y * (i - f));
      if (cur == INFLL)
        ok = false;
      else
        ans += cur;
      pr = v;
    }
  }
  if (ok)
    printf("%lld\n", ans);
  else
    printf("-1\n");
  return 0;
}

### Prompt

Create a solution in Cpp for 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>
const long long INFLL = ~((long long)1 << 63);
const int N = 200000;
int n, m, x, y, k, a[N + 1];
int main() {
  scanf("%d%d%d%d%d", &n, &m, &x, &k, &y);
  for (int i = 0; i < n; ++i) scanf("%d", a + i);
  int pr = -1, v;
  bool ok = true;
  long long ans = 0;
  for (int i = 0, j = 0; j <= m && ok; ++i, ++j) {
    if (j < m)
      scanf("%d", &v);
    else
      v = 0;
    int f = i, mx = 0;
    while (i <= n && a[i] != v) {
      mx = std::max(mx, a[i]);
      ++i;
    }
    if (i == n + 1)
      ok = false;
    else {
      long long cur = INFLL;
      if (i - f >= k)
        cur = std::min(
            (long long)x * ((i - f) / k) + (long long)y * ((i - f) % k),
            x + (long long)y * (i - f - k));
      if (mx < std::max(pr, v)) cur = std::min(cur, (long long)y * (i - f));
      if (cur == INFLL)
        ok = false;
      else
        ans += cur;
      pr = v;
    }
  }
  if (ok)
    printf("%lld\n", ans);
  else
    printf("-1\n");
  return 0;
}

```