#include <bits/stdc++.h>
using namespace std;
const int N = 200009;
int n, m, a[N], b[N], k, x, y, z, flag;
long long ans, sum, res;
int main() {
  scanf("%d%d", &n, &m);
  scanf("%d%d%d", &x, &k, &y);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= m; ++i) scanf("%d", &b[i]);
  int len = 0, now = 1, maxn = 0;
  for (int i = 1; i <= m + 1; ++i) {
    while (a[now] != b[i] && now <= n) ++len, maxn = max(maxn, a[now++]);
    if (now == n + 2) {
      puts("-1");
      return 0;
    }
    if (maxn > a[now - len - 1] && maxn > a[now]) {
      if (len < k) {
        puts("-1");
        return 0;
      } else {
        len -= k;
        ans += x;
      }
    }
    if (len > 0) {
      res = 1e16;
      for (int j = 0; j * k <= len; ++j)
        res = min(res, 1LL * j * x + 1LL * y * (len - j * k));
      ans += res;
    }
    ++now, len = maxn = 0;
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
const int N = 200009;
int n, m, a[N], b[N], k, x, y, z, flag;
long long ans, sum, res;
int main() {
  scanf("%d%d", &n, &m);
  scanf("%d%d%d", &x, &k, &y);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= m; ++i) scanf("%d", &b[i]);
  int len = 0, now = 1, maxn = 0;
  for (int i = 1; i <= m + 1; ++i) {
    while (a[now] != b[i] && now <= n) ++len, maxn = max(maxn, a[now++]);
    if (now == n + 2) {
      puts("-1");
      return 0;
    }
    if (maxn > a[now - len - 1] && maxn > a[now]) {
      if (len < k) {
        puts("-1");
        return 0;
      } else {
        len -= k;
        ans += x;
      }
    }
    if (len > 0) {
      res = 1e16;
      for (int j = 0; j * k <= len; ++j)
        res = min(res, 1LL * j * x + 1LL * y * (len - j * k));
      ans += res;
    }
    ++now, len = maxn = 0;
  }
  printf("%lld\n", ans);
  return 0;
}

```