#include <bits/stdc++.h>
using namespace std;
long long a[200005], pos[200005];
long long n, m, x, k, y, ans = 0, flag = 1;
long long calc(long long i, long long j) {
  long long mana = 0, Max = -10e17;
  for (long long p = i + 1; p < j; ++p) {
    Max = max(Max, a[p]);
  }
  if (j - i - 1 < k) {
    if (Max > max(a[i], a[j])) {
      flag = 0;
    }
    mana = y * (j - i - 1);
  } else {
    if (x < y * k) {
      mana = ((j - i - 1) / k) * x + ((j - i - 1) % k) * y;
    } else {
      if (Max > max(a[i], a[j])) {
        mana = x + (j - i - 1 - k) * y;
      } else {
        mana = (j - i - 1) * y;
      }
    }
  }
  return mana;
}
int main() {
  long long i, prev_pos = 0;
  scanf("%lld%lld%lld%lld%lld", &n, &m, &x, &k, &y);
  for (i = 1; i <= n; ++i) {
    scanf("%lld", &a[i]);
    pos[a[i]] = i;
  }
  a[n + 1] = 0;
  for (i = 1; i <= m; ++i) {
    long long val;
    scanf("%lld", &val);
    if (pos[val] < prev_pos) {
      flag = 0;
    }
    ans += calc(prev_pos, pos[val]);
    prev_pos = pos[val];
  }
  ans += calc(prev_pos, n + 1);
  if (flag) {
    printf("%lld", ans);
  } else {
    printf("-1");
  }
  return 0;
}

### Prompt

Please formulate 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 a[200005], pos[200005];
long long n, m, x, k, y, ans = 0, flag = 1;
long long calc(long long i, long long j) {
  long long mana = 0, Max = -10e17;
  for (long long p = i + 1; p < j; ++p) {
    Max = max(Max, a[p]);
  }
  if (j - i - 1 < k) {
    if (Max > max(a[i], a[j])) {
      flag = 0;
    }
    mana = y * (j - i - 1);
  } else {
    if (x < y * k) {
      mana = ((j - i - 1) / k) * x + ((j - i - 1) % k) * y;
    } else {
      if (Max > max(a[i], a[j])) {
        mana = x + (j - i - 1 - k) * y;
      } else {
        mana = (j - i - 1) * y;
      }
    }
  }
  return mana;
}
int main() {
  long long i, prev_pos = 0;
  scanf("%lld%lld%lld%lld%lld", &n, &m, &x, &k, &y);
  for (i = 1; i <= n; ++i) {
    scanf("%lld", &a[i]);
    pos[a[i]] = i;
  }
  a[n + 1] = 0;
  for (i = 1; i <= m; ++i) {
    long long val;
    scanf("%lld", &val);
    if (pos[val] < prev_pos) {
      flag = 0;
    }
    ans += calc(prev_pos, pos[val]);
    prev_pos = pos[val];
  }
  ans += calc(prev_pos, n + 1);
  if (flag) {
    printf("%lld", ans);
  } else {
    printf("-1");
  }
  return 0;
}

```