#include <bits/stdc++.h>
#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
using namespace std;
const long long N = 5e5 + 3;
const long long mod = 998244353;
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int INFi = 0x7f7f7f7f;
long long test = 1, n, m, x, y, k, b[N], a[N];
vector<long long> v;
long long ch(int l, int r) {
  long long mx = max(a[l], a[r]), res = 0;
  set<long long> s1, s2;
  for (int i = l + 1; i < r; i++) {
    if (a[i] > mx) s1.insert(i);
    s2.insert(i);
  }
  if (s2.size() < k and s1.size()) {
    cout << -1 << "\n";
    exit(0);
  }
  if (k * x > y) {
    res = (s2.size() / k) * y + (s2.size() % k) * x;
  } else {
    if (s1.size())
      res = y + (s2.size() - k) * x;
    else
      res = s2.size() * x;
  }
  return res;
}
void solve() {
  cin >> n >> m >> y >> k >> x;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= m; i++) cin >> b[i];
  a[0] = a[n + 1] = b[0] = b[m + 1] = 0;
  int j = 1;
  v.push_back(0);
  for (int i = 1; i <= n; i++)
    if (a[i] == b[j]) j++, v.push_back(i);
  v.push_back(n + 1);
  if (j != m + 1) {
    cout << -1 << "\n";
    return;
  }
  long long ans = 0;
  for (int i = 1; i <= m + 1; i++) {
    ans += ch(v[i - 1], v[i]);
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout << setprecision(15);
  ;
  for (int i = 1; i <= test; i++) {
    solve();
  }
}

### Prompt

Please create a solution in Cpp 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>
#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
using namespace std;
const long long N = 5e5 + 3;
const long long mod = 998244353;
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int INFi = 0x7f7f7f7f;
long long test = 1, n, m, x, y, k, b[N], a[N];
vector<long long> v;
long long ch(int l, int r) {
  long long mx = max(a[l], a[r]), res = 0;
  set<long long> s1, s2;
  for (int i = l + 1; i < r; i++) {
    if (a[i] > mx) s1.insert(i);
    s2.insert(i);
  }
  if (s2.size() < k and s1.size()) {
    cout << -1 << "\n";
    exit(0);
  }
  if (k * x > y) {
    res = (s2.size() / k) * y + (s2.size() % k) * x;
  } else {
    if (s1.size())
      res = y + (s2.size() - k) * x;
    else
      res = s2.size() * x;
  }
  return res;
}
void solve() {
  cin >> n >> m >> y >> k >> x;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= m; i++) cin >> b[i];
  a[0] = a[n + 1] = b[0] = b[m + 1] = 0;
  int j = 1;
  v.push_back(0);
  for (int i = 1; i <= n; i++)
    if (a[i] == b[j]) j++, v.push_back(i);
  v.push_back(n + 1);
  if (j != m + 1) {
    cout << -1 << "\n";
    return;
  }
  long long ans = 0;
  for (int i = 1; i <= m + 1; i++) {
    ans += ch(v[i - 1], v[i]);
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout << setprecision(15);
  ;
  for (int i = 1; i <= test; i++) {
    solve();
  }
}

```