#include <bits/stdc++.h>
using namespace std;
void my_dbg() { cout << endl; }
template <typename Arg, typename... Args>
void my_dbg(Arg A, Args... B) {
  cout << ' ' << A;
  my_dbg(B...);
}
using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;
const int inf = INT_MAX;
const int ninf = INT_MIN;
const int mod = 1e9 + 7;
const int N = 2e5 + 2;
vi a(N), b(N);
int n, m;
ll x, k, y;
bool perform(int l, int r, ll& res) {
  if (l > r) return 1;
  int len = r - l + 1;
  int toughest_opponent = *max_element(a.begin() + l, a.begin() + r + 1);
  bool ultimate_warrior = 0;
  if (l > 0 and a[l - 1] > toughest_opponent) ultimate_warrior = 1;
  if (r + 1 < n and a[r + 1] > toughest_opponent) ultimate_warrior = 1;
  if (len < k and !ultimate_warrior) return 0;
  int need = len % k;
  res += need * y;
  len -= need;
  if (y * k >= x) {
    res += len / k * x;
  } else if (ultimate_warrior) {
    res += len * y;
  } else {
    res += (len - k) * y + x;
  }
  return 1;
}
void solve() {
  scanf("%d", &n);
  scanf("%d", &m);
  scanf("%lld%lld%lld", &x, &k, &y);
  for (int i = (int)(0); i < (int)(n); i++) scanf("%d", &a[i]);
  for (int i = (int)(0); i < (int)(m); i++) scanf("%d", &b[i]);
  ll res = 0;
  int i = 0, j = 0, lst = -1;
  while (j < m) {
    while (i < n and b[j] != a[i]) i++;
    if (i == n) {
      printf("%d\n", (int)(-1));
      exit(0);
    }
    if (!perform(lst + 1, i - 1, res)) {
      printf("%d\n", (int)(-1));
      exit(0);
    }
    lst = i;
    j++;
  }
  if (!perform(lst + 1, n - 1, res)) {
    printf("%d\n", (int)(-1));
    exit(0);
  }
  printf("%lld\n", res);
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Your task is to create 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;
void my_dbg() { cout << endl; }
template <typename Arg, typename... Args>
void my_dbg(Arg A, Args... B) {
  cout << ' ' << A;
  my_dbg(B...);
}
using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;
const int inf = INT_MAX;
const int ninf = INT_MIN;
const int mod = 1e9 + 7;
const int N = 2e5 + 2;
vi a(N), b(N);
int n, m;
ll x, k, y;
bool perform(int l, int r, ll& res) {
  if (l > r) return 1;
  int len = r - l + 1;
  int toughest_opponent = *max_element(a.begin() + l, a.begin() + r + 1);
  bool ultimate_warrior = 0;
  if (l > 0 and a[l - 1] > toughest_opponent) ultimate_warrior = 1;
  if (r + 1 < n and a[r + 1] > toughest_opponent) ultimate_warrior = 1;
  if (len < k and !ultimate_warrior) return 0;
  int need = len % k;
  res += need * y;
  len -= need;
  if (y * k >= x) {
    res += len / k * x;
  } else if (ultimate_warrior) {
    res += len * y;
  } else {
    res += (len - k) * y + x;
  }
  return 1;
}
void solve() {
  scanf("%d", &n);
  scanf("%d", &m);
  scanf("%lld%lld%lld", &x, &k, &y);
  for (int i = (int)(0); i < (int)(n); i++) scanf("%d", &a[i]);
  for (int i = (int)(0); i < (int)(m); i++) scanf("%d", &b[i]);
  ll res = 0;
  int i = 0, j = 0, lst = -1;
  while (j < m) {
    while (i < n and b[j] != a[i]) i++;
    if (i == n) {
      printf("%d\n", (int)(-1));
      exit(0);
    }
    if (!perform(lst + 1, i - 1, res)) {
      printf("%d\n", (int)(-1));
      exit(0);
    }
    lst = i;
    j++;
  }
  if (!perform(lst + 1, n - 1, res)) {
    printf("%d\n", (int)(-1));
    exit(0);
  }
  printf("%lld\n", res);
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```