#include <bits/stdc++.h>
using namespace std;
long long inline mo(long long a) { return a % (long long)(1e9 + 7); }
long long po(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y >>= 1;
    x = (x * x) % p;
  }
  return res % p;
}
template <typename T>
void inline enter(vector<T> &a) {
  for (T &e : a) cin >> e;
}
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) 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...);
}
void JUDGE() {}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m;
  cin >> n >> m;
  long long cost_fire, range, cost_ber;
  cin >> cost_fire >> range >> cost_ber;
  vector<long long> a(n), b(m);
  enter(a);
  enter(b);
  b.push_back(-(long long)(1e18 + 5));
  a.push_back(-(long long)(1e18 + 5));
  n++;
  m++;
  long long pos = 1, j = 0;
  vector<vector<long long>> spaces;
  for (long long i = 0; i < m; i++) {
    vector<long long> space;
    while (j < n && a[j] != b[i]) {
      space.push_back(a[j]);
      j++;
    }
    if (j >= n) {
      pos = 0;
      break;
    }
    j++;
    spaces.push_back(space);
  }
  if (!pos) {
    cout << -1 << "\n";
    return 0;
  }
  long long si = spaces.size(), ans = 0;
  for (long long i = 0; i < si; i++) {
    long long cur_si = spaces[i].size();
    if (!cur_si) continue;
    vector<long long> space = spaces[i];
    long long left = -(long long)(1e18 + 5), right = -(long long)(1e18 + 5);
    if (i) left = b[i - 1];
    right = b[i];
    if (cur_si < range) {
      if (*max_element(begin(space), end(space)) > max(left, right)) {
        pos = 0;
        break;
      } else {
        ans += cost_ber * cur_si;
        continue;
      }
    }
    if (range * cost_ber <= cost_fire) {
      long long great = 0;
      for (long long j = 0; j < cur_si; j++)
        great += (space[j] > max(left, right));
      if (!great) {
        ans += cost_ber * cur_si;
        continue;
      }
      ans += cost_fire + (cur_si - range) * cost_ber;
    } else {
      long long rem = cur_si % range;
      ans += rem * cost_ber + (cur_si / range) * cost_fire;
    }
  }
  if (!pos) {
    cout << -1 << "\n";
    return 0;
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Generate 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;
long long inline mo(long long a) { return a % (long long)(1e9 + 7); }
long long po(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y >>= 1;
    x = (x * x) % p;
  }
  return res % p;
}
template <typename T>
void inline enter(vector<T> &a) {
  for (T &e : a) cin >> e;
}
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) 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...);
}
void JUDGE() {}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m;
  cin >> n >> m;
  long long cost_fire, range, cost_ber;
  cin >> cost_fire >> range >> cost_ber;
  vector<long long> a(n), b(m);
  enter(a);
  enter(b);
  b.push_back(-(long long)(1e18 + 5));
  a.push_back(-(long long)(1e18 + 5));
  n++;
  m++;
  long long pos = 1, j = 0;
  vector<vector<long long>> spaces;
  for (long long i = 0; i < m; i++) {
    vector<long long> space;
    while (j < n && a[j] != b[i]) {
      space.push_back(a[j]);
      j++;
    }
    if (j >= n) {
      pos = 0;
      break;
    }
    j++;
    spaces.push_back(space);
  }
  if (!pos) {
    cout << -1 << "\n";
    return 0;
  }
  long long si = spaces.size(), ans = 0;
  for (long long i = 0; i < si; i++) {
    long long cur_si = spaces[i].size();
    if (!cur_si) continue;
    vector<long long> space = spaces[i];
    long long left = -(long long)(1e18 + 5), right = -(long long)(1e18 + 5);
    if (i) left = b[i - 1];
    right = b[i];
    if (cur_si < range) {
      if (*max_element(begin(space), end(space)) > max(left, right)) {
        pos = 0;
        break;
      } else {
        ans += cost_ber * cur_si;
        continue;
      }
    }
    if (range * cost_ber <= cost_fire) {
      long long great = 0;
      for (long long j = 0; j < cur_si; j++)
        great += (space[j] > max(left, right));
      if (!great) {
        ans += cost_ber * cur_si;
        continue;
      }
      ans += cost_fire + (cur_si - range) * cost_ber;
    } else {
      long long rem = cur_si % range;
      ans += rem * cost_ber + (cur_si / range) * cost_fire;
    }
  }
  if (!pos) {
    cout << -1 << "\n";
    return 0;
  }
  cout << ans << "\n";
  return 0;
}

```