#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1e9 + 7;
const long long int mod1 = 998244353;
int main() {
  long long int T;
  T = 1;
  while (T--) {
    long long int n, m, x, k, y, i, j, l;
    cin >> n >> m >> x >> k >> y;
    long long int a[n], b[m];
    for (i = 0; i < n; i++) cin >> a[i];
    for (i = 0; i < m; i++) cin >> b[i];
    vector<pair<long long int, long long int>> v;
    j = 0, l = 0;
    for (i = 0; i < n;) {
      if (a[i] == b[j]) {
        j++;
        i++;
        continue;
      }
      l = i;
      long long int flag1 = 0;
      while ((l < n) && (a[l] != b[j])) {
        if (((j > 0) and (a[l] < b[j - 1])) || ((j != m) && (a[l] < b[j])))
          l++;
        else
          flag1 = 1, l++;
      }
      long long int range = l - i;
      i = l;
      j++;
      i++;
      v.push_back({range, flag1});
    }
    if (j < m) {
      cout << "-1\n";
      continue;
    }
    if (v.empty()) {
      cout << 0 << "\n";
      continue;
    }
    bool flag = true;
    for (i = 0; i < v.size(); i++) {
      if (v[i].first < (v[i].second * k)) flag = false;
    }
    if (flag == false) {
      cout << "-1\n";
      continue;
    }
    long long int ans = 0;
    for (i = 0; i < v.size(); i++) {
      ans += (v[i].second * x);
      v[i].first -= (v[i].second * k);
      ans += (min((v[i].first / k) * x, (v[i].first / k) * k * y));
      ans += (v[i].first % k) * y;
    }
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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 long long int mod = 1e9 + 7;
const long long int mod1 = 998244353;
int main() {
  long long int T;
  T = 1;
  while (T--) {
    long long int n, m, x, k, y, i, j, l;
    cin >> n >> m >> x >> k >> y;
    long long int a[n], b[m];
    for (i = 0; i < n; i++) cin >> a[i];
    for (i = 0; i < m; i++) cin >> b[i];
    vector<pair<long long int, long long int>> v;
    j = 0, l = 0;
    for (i = 0; i < n;) {
      if (a[i] == b[j]) {
        j++;
        i++;
        continue;
      }
      l = i;
      long long int flag1 = 0;
      while ((l < n) && (a[l] != b[j])) {
        if (((j > 0) and (a[l] < b[j - 1])) || ((j != m) && (a[l] < b[j])))
          l++;
        else
          flag1 = 1, l++;
      }
      long long int range = l - i;
      i = l;
      j++;
      i++;
      v.push_back({range, flag1});
    }
    if (j < m) {
      cout << "-1\n";
      continue;
    }
    if (v.empty()) {
      cout << 0 << "\n";
      continue;
    }
    bool flag = true;
    for (i = 0; i < v.size(); i++) {
      if (v[i].first < (v[i].second * k)) flag = false;
    }
    if (flag == false) {
      cout << "-1\n";
      continue;
    }
    long long int ans = 0;
    for (i = 0; i < v.size(); i++) {
      ans += (v[i].second * x);
      v[i].first -= (v[i].second * k);
      ans += (min((v[i].first / k) * x, (v[i].first / k) * k * y));
      ans += (v[i].first % k) * y;
    }
    cout << ans << "\n";
  }
  return 0;
}

```