#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <typename T>
inline bool chkmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T>
inline bool chkmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int N = 2e5;
int n, m, x, k, y, a[N], b[N];
ll res;
int process(vector<int> &v, int val1, int val2) {
  if ((int)(v.size()) == 0) return 0;
  sort(v.begin(), v.end());
  if (k == 1) {
    if (x <= y) {
      res += 1ll * x * (int)(v.size());
    } else {
      res += 1ll * y * ((int)(v.size()) - 1);
      if (v.back() > val1 && v.back() > val2)
        res += x;
      else
        res += y;
    }
  } else {
    if ((int)(v.size()) < k) {
      res += 1ll * y * ((int)(v.size()) - 1);
      if (val1 > v.back() || val2 > v.back())
        res += y;
      else
        return -1;
    } else {
      int f = 0;
      if (v.back() > val1 && v.back() > val2) f = 1;
      ll add = 1e18;
      while (k * f <= (int)(v.size())) {
        int yy = (int)(v.size()) - k * f;
        chkmin(add, 1ll * f * x + 1ll * yy * y);
        ++f;
      }
      if (add == ll(1e18)) return -1;
      res += add;
    }
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m >> x >> k >> y;
  for (int i = (int)(0); i < (int)(n); ++i) {
    cin >> a[i];
  }
  for (int i = (int)(0); i < (int)(m); ++i) {
    cin >> b[i];
  }
  int id = 0;
  vector<int> v;
  for (int i = (int)(0); i < (int)(n); ++i) {
    if (a[i] == b[id]) {
      if (process(v, b[id], (id == 0 ? -1 : b[id - 1])) == -1) {
        cout << -1 << "\n";
        return 0;
      }
      id++;
      v.clear();
      continue;
    }
    v.push_back(a[i]);
  }
  if (id != m || process(v, b[m - 1], -1) == -1) {
    cout << -1 << "\n";
  } else
    cout << res << "\n";
  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;
using ll = long long;
template <typename T>
inline bool chkmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T>
inline bool chkmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int N = 2e5;
int n, m, x, k, y, a[N], b[N];
ll res;
int process(vector<int> &v, int val1, int val2) {
  if ((int)(v.size()) == 0) return 0;
  sort(v.begin(), v.end());
  if (k == 1) {
    if (x <= y) {
      res += 1ll * x * (int)(v.size());
    } else {
      res += 1ll * y * ((int)(v.size()) - 1);
      if (v.back() > val1 && v.back() > val2)
        res += x;
      else
        res += y;
    }
  } else {
    if ((int)(v.size()) < k) {
      res += 1ll * y * ((int)(v.size()) - 1);
      if (val1 > v.back() || val2 > v.back())
        res += y;
      else
        return -1;
    } else {
      int f = 0;
      if (v.back() > val1 && v.back() > val2) f = 1;
      ll add = 1e18;
      while (k * f <= (int)(v.size())) {
        int yy = (int)(v.size()) - k * f;
        chkmin(add, 1ll * f * x + 1ll * yy * y);
        ++f;
      }
      if (add == ll(1e18)) return -1;
      res += add;
    }
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m >> x >> k >> y;
  for (int i = (int)(0); i < (int)(n); ++i) {
    cin >> a[i];
  }
  for (int i = (int)(0); i < (int)(m); ++i) {
    cin >> b[i];
  }
  int id = 0;
  vector<int> v;
  for (int i = (int)(0); i < (int)(n); ++i) {
    if (a[i] == b[id]) {
      if (process(v, b[id], (id == 0 ? -1 : b[id - 1])) == -1) {
        cout << -1 << "\n";
        return 0;
      }
      id++;
      v.clear();
      continue;
    }
    v.push_back(a[i]);
  }
  if (id != m || process(v, b[m - 1], -1) == -1) {
    cout << -1 << "\n";
  } else
    cout << res << "\n";
  return 0;
}

```