#include <bits/stdc++.h>
using namespace std;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << '{';
  string sep;
  for (const auto &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename T, size_t size>
ostream &operator<<(ostream &os, const array<T, size> &arr) {
  os << '{';
  string sep;
  for (const auto &x : arr) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
const int64_t INF64 = int64_t(2e18) + 5;
int N, M, K;
int64_t X, Y;
int64_t solve(const vector<int> &A, int start, int end) {
  int count = end - start;
  if (count <= 0) return 0;
  int bigger = max(start <= 0 ? 0 : A[start - 1], end >= N ? 0 : A[end]);
  if (bigger > *max_element(A.begin() + start, A.begin() + end)) {
    return (count / K) * min(X, Y * K) + (count % K) * Y;
  } else {
    if (count < K) return INF64;
    return X + (count / K - 1) * min(X, Y * K) + (count % K) * Y;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> N >> M >> X >> K >> Y;
  vector<int> A(N), B(M);
  for (auto &a : A) cin >> a;
  for (auto &b : B) cin >> b;
  int64_t answer = 0;
  int position = 0;
  for (auto &b : B) {
    int start = position;
    while (position < N && A[position] != b) position++;
    if (position >= N) {
      cout << -1 << '\n';
      return 0;
    }
    assert(A[position] == b);
    answer += solve(A, start, position);
    position++;
    if (answer >= INF64) {
      cout << -1 << '\n';
      return 0;
    }
  }
  if (position < N) answer += solve(A, position, N);
  cout << answer << '\n';
}

### Prompt

Please create a solution in Cpp 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;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << '{';
  string sep;
  for (const auto &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename T, size_t size>
ostream &operator<<(ostream &os, const array<T, size> &arr) {
  os << '{';
  string sep;
  for (const auto &x : arr) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
const int64_t INF64 = int64_t(2e18) + 5;
int N, M, K;
int64_t X, Y;
int64_t solve(const vector<int> &A, int start, int end) {
  int count = end - start;
  if (count <= 0) return 0;
  int bigger = max(start <= 0 ? 0 : A[start - 1], end >= N ? 0 : A[end]);
  if (bigger > *max_element(A.begin() + start, A.begin() + end)) {
    return (count / K) * min(X, Y * K) + (count % K) * Y;
  } else {
    if (count < K) return INF64;
    return X + (count / K - 1) * min(X, Y * K) + (count % K) * Y;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> N >> M >> X >> K >> Y;
  vector<int> A(N), B(M);
  for (auto &a : A) cin >> a;
  for (auto &b : B) cin >> b;
  int64_t answer = 0;
  int position = 0;
  for (auto &b : B) {
    int start = position;
    while (position < N && A[position] != b) position++;
    if (position >= N) {
      cout << -1 << '\n';
      return 0;
    }
    assert(A[position] == b);
    answer += solve(A, start, position);
    position++;
    if (answer >= INF64) {
      cout << -1 << '\n';
      return 0;
    }
  }
  if (position < N) answer += solve(A, position, N);
  cout << answer << '\n';
}

```