#include <bits/stdc++.h>
using namespace std;
const long long max_n = 2e5 + 5;
long long n, m, x, y, k;
vector<long long> a, b;
long long cost(pair<long long, long long> ivl) {
  long long val = 0;
  if (ivl.first > 0) val = max(val, a[ivl.first - 1]);
  if (ivl.second < n - 1) val = max(val, a[ivl.second + 1]);
  long long sz = ivl.second - ivl.first + 1, mx = 0;
  for (long long i = ivl.first;
       (1 > 0 and i < ivl.second + 1) or (1 < 0 and i > ivl.second + 1); i += 1)
    mx = max(mx, a[i]);
  if (mx > val) {
    if (sz < k) return -1;
    return min(((sz / k) * x) + ((sz % k) * y), x + ((sz - k) * y));
  } else
    return min(((sz / k) * x) + ((sz % k) * y), sz * y);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m;
  cin >> x >> k >> y;
  a.resize(n);
  for (long long z = 0; (1 > 0 and z < n) or (1 < 0 and z > n); z += 1)
    cin >> a[z];
  ;
  b.resize(m);
  for (long long z = 0; (1 > 0 and z < m) or (1 < 0 and z > m); z += 1)
    cin >> b[z];
  ;
  vector<long long> pos(n + 1, -1);
  for (long long i = 0; (1 > 0 and i < n) or (1 < 0 and i > n); i += 1)
    pos[a[i]] = i;
  bool flag = true;
  for (long long i = 1; (1 > 0 and i < m) or (1 < 0 and i > m); i += 1)
    if (pos[b[i - 1]] > pos[b[i]]) flag = false;
  if (!flag) {
    cout << -1 << endl;
    return 0;
  }
  vector<pair<long long, long long> > ivl;
  if (pos[b[0]] > 0) ivl.push_back({0, pos[b[0]] - 1});
  for (long long i = 1; (1 > 0 and i < m) or (1 < 0 and i > m); i += 1)
    if (pos[b[i]] - pos[b[i - 1]] > 1)
      ivl.push_back({pos[b[i - 1]] + 1, pos[b[i]] - 1});
  if (pos[b[m - 1]] < n - 1) ivl.push_back({pos[b[m - 1]] + 1, n - 1});
  long long ans = 0;
  for (auto p : ivl) {
    long long val = cost(p);
    if (val == -1) {
      cout << -1 << endl;
      return 0;
    }
    ans += val;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 max_n = 2e5 + 5;
long long n, m, x, y, k;
vector<long long> a, b;
long long cost(pair<long long, long long> ivl) {
  long long val = 0;
  if (ivl.first > 0) val = max(val, a[ivl.first - 1]);
  if (ivl.second < n - 1) val = max(val, a[ivl.second + 1]);
  long long sz = ivl.second - ivl.first + 1, mx = 0;
  for (long long i = ivl.first;
       (1 > 0 and i < ivl.second + 1) or (1 < 0 and i > ivl.second + 1); i += 1)
    mx = max(mx, a[i]);
  if (mx > val) {
    if (sz < k) return -1;
    return min(((sz / k) * x) + ((sz % k) * y), x + ((sz - k) * y));
  } else
    return min(((sz / k) * x) + ((sz % k) * y), sz * y);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m;
  cin >> x >> k >> y;
  a.resize(n);
  for (long long z = 0; (1 > 0 and z < n) or (1 < 0 and z > n); z += 1)
    cin >> a[z];
  ;
  b.resize(m);
  for (long long z = 0; (1 > 0 and z < m) or (1 < 0 and z > m); z += 1)
    cin >> b[z];
  ;
  vector<long long> pos(n + 1, -1);
  for (long long i = 0; (1 > 0 and i < n) or (1 < 0 and i > n); i += 1)
    pos[a[i]] = i;
  bool flag = true;
  for (long long i = 1; (1 > 0 and i < m) or (1 < 0 and i > m); i += 1)
    if (pos[b[i - 1]] > pos[b[i]]) flag = false;
  if (!flag) {
    cout << -1 << endl;
    return 0;
  }
  vector<pair<long long, long long> > ivl;
  if (pos[b[0]] > 0) ivl.push_back({0, pos[b[0]] - 1});
  for (long long i = 1; (1 > 0 and i < m) or (1 < 0 and i > m); i += 1)
    if (pos[b[i]] - pos[b[i - 1]] > 1)
      ivl.push_back({pos[b[i - 1]] + 1, pos[b[i]] - 1});
  if (pos[b[m - 1]] < n - 1) ivl.push_back({pos[b[m - 1]] + 1, n - 1});
  long long ans = 0;
  for (auto p : ivl) {
    long long val = cost(p);
    if (val == -1) {
      cout << -1 << endl;
      return 0;
    }
    ans += val;
  }
  cout << ans << endl;
  return 0;
}

```