#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  long long x, k, y;
  scanf("%lld%lld%lld", &x, &k, &y);
  int a[n];
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int b[m + 1];
  for (int i = 0; i < m; i++) scanf("%d", &b[i]);
  b[m] = -1;
  long long res = 0;
  int pos = 0, last = -1;
  for (int i = 0; i <= m; i++) {
    int t = 0, start = pos, p = b[i];
    while (pos < n && a[pos] != p) {
      t = max(t, a[pos++]);
    }
    if (pos == n && i < m) {
      res = -1;
      break;
    }
    int cnt = pos - start;
    if (cnt > 0) {
      if (cnt < k) {
        if (t > last && t > p) {
          res = -1;
          break;
        }
        res += cnt * y;
      } else {
        if (x < k * y) {
          res += (cnt / k) * x + (cnt % k) * y;
        } else {
          if (t > last && t > p) {
            res += x + (cnt - k) * y;
          } else {
            res += cnt * y;
          }
        }
      }
    }
    last = p;
    pos++;
  }
  printf("%lld\r\n", res);
}

### 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;
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  long long x, k, y;
  scanf("%lld%lld%lld", &x, &k, &y);
  int a[n];
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int b[m + 1];
  for (int i = 0; i < m; i++) scanf("%d", &b[i]);
  b[m] = -1;
  long long res = 0;
  int pos = 0, last = -1;
  for (int i = 0; i <= m; i++) {
    int t = 0, start = pos, p = b[i];
    while (pos < n && a[pos] != p) {
      t = max(t, a[pos++]);
    }
    if (pos == n && i < m) {
      res = -1;
      break;
    }
    int cnt = pos - start;
    if (cnt > 0) {
      if (cnt < k) {
        if (t > last && t > p) {
          res = -1;
          break;
        }
        res += cnt * y;
      } else {
        if (x < k * y) {
          res += (cnt / k) * x + (cnt % k) * y;
        } else {
          if (t > last && t > p) {
            res += x + (cnt - k) * y;
          } else {
            res += cnt * y;
          }
        }
      }
    }
    last = p;
    pos++;
  }
  printf("%lld\r\n", res);
}

```