#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
template <typename T>
bool mmax(T &m, const T q) {
  if (m < q) {
    m = q;
    return true;
  } else
    return false;
}
template <typename T>
bool mmin(T &m, const T q) {
  if (m > q) {
    m = q;
    return true;
  } else
    return false;
}
void __print(long long first) { cerr << first; }
void __print(long first) { cerr << first; }
void __print(unsigned first) { cerr << first; }
void __print(unsigned long first) { cerr << first; }
void __print(unsigned long long first) { cerr << first; }
void __print(float first) { cerr << first; }
void __print(double first) { cerr << first; }
void __print(long double first) { cerr << first; }
void __print(char first) { cerr << '\'' << first << '\''; }
void __print(const char *first) { cerr << '\"' << first << '\"'; }
void __print(const string &first) { cerr << '\"' << first << '\"'; }
void __print(bool first) { cerr << (first ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &first) {
  cerr << '{';
  __print(first.first);
  cerr << ',';
  __print(first.second);
  cerr << '}';
}
template <typename T>
void __print(const T &first) {
  long long f = 0;
  cerr << '{';
  for (auto &i : first) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
std::mt19937 rng(
    (long long)std::chrono::steady_clock::now().time_since_epoch().count());
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(12);
  long long n, m, first, k, second;
  cin >> n >> m >> first >> k >> second;
  long long a[n + 1], b[m + 1];
  map<long long, long long> pos;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    pos[a[i]] = i;
  }
  for (long long i = 1; i <= m; i++) cin >> b[i];
  a[0] = -1;
  a[n + 1] = -2;
  pos[-1] = 0;
  pos[-2] = n + 1;
  b[0] = -1;
  b[m + 1] = -2;
  long long ans = 0;
  for (long long i = 1; i <= m + 1; i++) {
    long long l = pos[b[i]] - pos[b[i - 1]] - 1;
    if (l < 0) {
      cout << -1;
      exit(0);
    }
    if (l == 0) continue;
    long long cost = 0;
    bool has = false;
    long long c = max(b[i], b[i - 1]);
    for (long long j = pos[b[i - 1]] + 1; j < pos[b[i]]; j++) {
      long long d = a[j];
      if (d > c) {
        has = true;
      }
    }
    if (has) {
      cost += first;
      l -= k;
    }
    if (l < 0) {
      cout << -1;
      exit(0);
    }
    long long op = (l / k) * k;
    cost += min(l * second, (l / k) * first + (l - op) * second);
    ans += cost;
  }
  cout << ans << "\n";
  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;
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
template <typename T>
bool mmax(T &m, const T q) {
  if (m < q) {
    m = q;
    return true;
  } else
    return false;
}
template <typename T>
bool mmin(T &m, const T q) {
  if (m > q) {
    m = q;
    return true;
  } else
    return false;
}
void __print(long long first) { cerr << first; }
void __print(long first) { cerr << first; }
void __print(unsigned first) { cerr << first; }
void __print(unsigned long first) { cerr << first; }
void __print(unsigned long long first) { cerr << first; }
void __print(float first) { cerr << first; }
void __print(double first) { cerr << first; }
void __print(long double first) { cerr << first; }
void __print(char first) { cerr << '\'' << first << '\''; }
void __print(const char *first) { cerr << '\"' << first << '\"'; }
void __print(const string &first) { cerr << '\"' << first << '\"'; }
void __print(bool first) { cerr << (first ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &first) {
  cerr << '{';
  __print(first.first);
  cerr << ',';
  __print(first.second);
  cerr << '}';
}
template <typename T>
void __print(const T &first) {
  long long f = 0;
  cerr << '{';
  for (auto &i : first) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
std::mt19937 rng(
    (long long)std::chrono::steady_clock::now().time_since_epoch().count());
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(12);
  long long n, m, first, k, second;
  cin >> n >> m >> first >> k >> second;
  long long a[n + 1], b[m + 1];
  map<long long, long long> pos;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    pos[a[i]] = i;
  }
  for (long long i = 1; i <= m; i++) cin >> b[i];
  a[0] = -1;
  a[n + 1] = -2;
  pos[-1] = 0;
  pos[-2] = n + 1;
  b[0] = -1;
  b[m + 1] = -2;
  long long ans = 0;
  for (long long i = 1; i <= m + 1; i++) {
    long long l = pos[b[i]] - pos[b[i - 1]] - 1;
    if (l < 0) {
      cout << -1;
      exit(0);
    }
    if (l == 0) continue;
    long long cost = 0;
    bool has = false;
    long long c = max(b[i], b[i - 1]);
    for (long long j = pos[b[i - 1]] + 1; j < pos[b[i]]; j++) {
      long long d = a[j];
      if (d > c) {
        has = true;
      }
    }
    if (has) {
      cost += first;
      l -= k;
    }
    if (l < 0) {
      cout << -1;
      exit(0);
    }
    long long op = (l / k) * k;
    cost += min(l * second, (l / k) * first + (l - op) * second);
    ans += cost;
  }
  cout << ans << "\n";
  return 0;
}

```