#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
long long A[N], B[N], bucket[N];
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  long long first, second, k;
  scanf("%lld %lld %lld", &first, &k, &second);
  for (int i = (1); i <= (n); ++i) scanf("%lld", &A[i]);
  for (int i = (1); i <= (m); ++i) scanf("%lld", &B[i]);
  int ptr = 0;
  long long mx = 0, cnt = 0;
  long long ans = 0;
  for (int i = (1); i <= (n); ++i) {
    if (ptr < m && A[i] == B[ptr + 1]) {
      ++ptr;
      bucket[ptr] = i;
      if (mx == 0) {
        mx = cnt = 0;
        continue;
      }
      if (cnt >= k) {
        long long add = (cnt / k) * first + (cnt % k) * second;
        if (mx < max(B[ptr], B[ptr - 1]))
          add = min(add, cnt * second);
        else
          add = min(add, first + (cnt - k) * second);
        ans += add;
      } else {
        if (mx < max(B[ptr], B[ptr - 1]))
          ans += cnt * second;
        else
          return !printf("-1");
      }
      mx = cnt = 0;
    } else {
      mx = max(mx, A[i]);
      ++cnt;
    }
  }
  if (ptr != m) return !printf("-1");
  if (cnt >= k) {
    long long add = (cnt / k) * first + (cnt % k) * second;
    if (mx < B[m])
      add = min(add, cnt * second);
    else
      add = min(add, first + (cnt - k) * second);
    ans += add;
  } else {
    if (mx < B[m])
      ans += cnt * second;
    else
      return !printf("-1");
  }
  printf("%lld", ans);
  return 0;
}

### Prompt

Your challenge is to write 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;
const int N = 200005;
long long A[N], B[N], bucket[N];
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  long long first, second, k;
  scanf("%lld %lld %lld", &first, &k, &second);
  for (int i = (1); i <= (n); ++i) scanf("%lld", &A[i]);
  for (int i = (1); i <= (m); ++i) scanf("%lld", &B[i]);
  int ptr = 0;
  long long mx = 0, cnt = 0;
  long long ans = 0;
  for (int i = (1); i <= (n); ++i) {
    if (ptr < m && A[i] == B[ptr + 1]) {
      ++ptr;
      bucket[ptr] = i;
      if (mx == 0) {
        mx = cnt = 0;
        continue;
      }
      if (cnt >= k) {
        long long add = (cnt / k) * first + (cnt % k) * second;
        if (mx < max(B[ptr], B[ptr - 1]))
          add = min(add, cnt * second);
        else
          add = min(add, first + (cnt - k) * second);
        ans += add;
      } else {
        if (mx < max(B[ptr], B[ptr - 1]))
          ans += cnt * second;
        else
          return !printf("-1");
      }
      mx = cnt = 0;
    } else {
      mx = max(mx, A[i]);
      ++cnt;
    }
  }
  if (ptr != m) return !printf("-1");
  if (cnt >= k) {
    long long add = (cnt / k) * first + (cnt % k) * second;
    if (mx < B[m])
      add = min(add, cnt * second);
    else
      add = min(add, first + (cnt - k) * second);
    ans += add;
  } else {
    if (mx < B[m])
      ans += cnt * second;
    else
      return !printf("-1");
  }
  printf("%lld", ans);
  return 0;
}

```