#include <bits/stdc++.h>
using namespace std;
const int MX = 2e5 + 5;
int n, m, a[MX], b[MX];
long long int x, k, y;
vector<int> pos;
long long int solve(int L, int R) {
  int num = R - L - 1;
  bool berserk;
  int tmp = 0;
  for (int i = L + 1; i < R; i++) tmp = max(tmp, a[i]);
  berserk = (tmp < a[L] || tmp < a[R]);
  if (num < k) {
    if (berserk)
      return num * y;
    else
      return -1;
  }
  if (k * y > x) {
    int fireball = num / k;
    num %= k;
    return x * fireball + y * num;
  }
  if (berserk)
    return num * y;
  else
    return x + (num - k) * y;
}
int main() {
  cin.tie(nullptr), ios::sync_with_stdio(false);
  cin >> n >> m >> x >> k >> y;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= m; i++) cin >> b[i];
  int p = 1;
  pos.push_back(0);
  for (int i = 1; i <= m; i++) {
    while (a[p] != b[i] && p <= n) p++;
    if (p > n) {
      cout << "-1";
      return 0;
    }
    pos.push_back(p);
  }
  pos.push_back(n + 1);
  long long int ans = 0;
  for (int i = 0; i < (int)pos.size() - 1; i++) {
    long long int tmp = solve(pos[i], pos[i + 1]);
    if (tmp == -1) {
      cout << "-1";
      return 0;
    }
    ans += tmp;
  }
  cout << ans;
}

### 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 MX = 2e5 + 5;
int n, m, a[MX], b[MX];
long long int x, k, y;
vector<int> pos;
long long int solve(int L, int R) {
  int num = R - L - 1;
  bool berserk;
  int tmp = 0;
  for (int i = L + 1; i < R; i++) tmp = max(tmp, a[i]);
  berserk = (tmp < a[L] || tmp < a[R]);
  if (num < k) {
    if (berserk)
      return num * y;
    else
      return -1;
  }
  if (k * y > x) {
    int fireball = num / k;
    num %= k;
    return x * fireball + y * num;
  }
  if (berserk)
    return num * y;
  else
    return x + (num - k) * y;
}
int main() {
  cin.tie(nullptr), ios::sync_with_stdio(false);
  cin >> n >> m >> x >> k >> y;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= m; i++) cin >> b[i];
  int p = 1;
  pos.push_back(0);
  for (int i = 1; i <= m; i++) {
    while (a[p] != b[i] && p <= n) p++;
    if (p > n) {
      cout << "-1";
      return 0;
    }
    pos.push_back(p);
  }
  pos.push_back(n + 1);
  long long int ans = 0;
  for (int i = 0; i < (int)pos.size() - 1; i++) {
    long long int tmp = solve(pos[i], pos[i + 1]);
    if (tmp == -1) {
      cout << "-1";
      return 0;
    }
    ans += tmp;
  }
  cout << ans;
}

```